Higher Order Linear Differential Equations III (Non-Constant Coefficients)
Higher Order Linear Differential Equations With Non-Constant Coefficients
Linear Differential Equation Of Order n Definition: An equation of the form \(a_n(x) \frac{d^n y}{d x^n}+a_{n-1}(x) \frac{d^{n-1} y}{d x^{n-1}}+\cdots+a_1(x) \frac{d y}{d x}\)\(+a_0(x) y=\mathrm{Q}(x)\) where \(a_0, a_P, \ldots, a_{n-1}, a_n\) and Q are continuous real functions in x defined on an interval \(I\) is called a linear differential equation of order n over the interval \(I\).
An equation of the form : \(\frac{d^2 y}{d x^2}+\mathrm{P}(x) \frac{d y}{d x}+\mathrm{Q}(x)=\mathrm{R}(x)\) where P(x), Q(x) and R(x) are real-valued functions of x defined on an interval \(I\), is called the linear equation of the second order with variable coefficients.
If P and Q are real constants, the linear equation can be solved by the methods discussed in the previous chapter. Otherwise, there is no general method known to solve the linear equation of the second order with variable coefficients. In this chapter, we discuss some methods which at times will yield a solution.
The linear equation of the second order with variable coefficients can be solved by the following methods.
- Change of the dependent variable, when part of the C. F. is known.
- Variation of parameters.
- Hereafter in this chapter P(x), Q(x), and R(x) are written as P, Q, and R.
Higher Order Linear Differential Equations 3 (Non-Constant Coefficients) General Solution Of \(\frac{d^2 y}{d x^2}+\mathrm{P} \frac{d y}{d x}+\mathrm{Q} y=\mathrm{R}\) By The Method Of Variation Of Parameters:
Given linear differential equation is \(\frac{d^2 y}{d x^2}+\mathrm{P} \frac{d y}{d x}+\mathrm{Qy}=\mathrm{R}\) ……..(1) where P and. Q are functions of x or real constants and R is only a function of x.
Its homogeneous equation corresponding to (1) is \(\frac{d^2 y}{d x^2}+\mathrm{P} \frac{d y}{d x}+\mathrm{Q} y=0\). Let \(y_c=c_1 u+c_2 v\)……(2) be the general solution of (2) where u and v are functions of x, and \(c_1, c_2\) are real constants hence it is the C. F. of (1).
y = \(c_1 u+c_2 v \text { satisfies }(2) \Rightarrow\left(c_1 u_2+c_2 v_2\right)+\mathrm{P}\left(c_1 u_1+c_2 v_1\right)+\mathrm{Q}\left(c_1 u+c_2 v\right)=0\)
⇒ \(c_1\left(u_2+\mathrm{P} u_1+\mathrm{Q} u\right)+c_2\left(v_2+\mathrm{P} v_1+\mathrm{Q} v\right)=0\)
⇒ \(u_2+\mathrm{P} u_1+\mathrm{Q} u=0 \ldots \text { (3) and } \Rightarrow v_2+\mathrm{P} v_1+\mathrm{Q} v=0 \ldots \text { (4) }\)
Let a particular integral \(y_p\) of (1) be \(y_p=\mathrm{A} u+\mathrm{B} v\)…….(5)
This is obtained from C.F. of (1) by replacing \(c_1\) and \(c_2\) with A and B respectively which are also some functions of x.
⇒ \(\frac{d y_p}{d x}=\mathrm{A} u_1+u \frac{d \mathrm{~A}}{d x}+\mathrm{B} v_1+v \frac{d \mathrm{~B}}{d x}=\left(\mathrm{A} u_1+\mathrm{B} v_1\right)+u \frac{d \mathrm{~A}}{d x}+v \frac{d \mathrm{~B}}{d x}\)
Choose A and B such that \(u \frac{d \mathrm{~A}}{d x}+v \frac{d \mathrm{~B}}{d x}=0\)…..(6)
Then \(\frac{d y_p}{d x}=\mathrm{A} u_1+\mathrm{B} v_1 \Rightarrow \frac{d^2 y_p}{d x^2}=\left(\mathrm{A} u_2+\mathrm{B} v_2\right)+u_1 \frac{d \mathrm{~A}}{d x}+v_1 \frac{d \mathrm{~B}}{d x}\)
And \(y_p=\mathrm{A} u+\mathrm{B} v\) Substituting these values in (1), we get: \((\mathrm{A} u_2+\mathrm{B} v_2+u_1 \frac{d \mathrm{~A}}{d x}+v_1 \frac{d \mathrm{~B}}{d x}\)
+ \(\mathrm{P}\left(\mathrm{A} u_1+\mathrm{B} v_1\right)+\mathrm{Q}(\mathrm{A} u+\mathrm{B} v)=\mathrm{R}.\)……(7)
⇒ \(\mathrm{A}\left(u_2+\mathrm{P} u_1+\mathrm{Q} u\right)+\mathrm{B}\left(v_2+\mathrm{P} v_1+\mathrm{Q} v\right)+\left(u_1 \frac{d \mathrm{~A}}{d x}+v_1 \frac{d \mathrm{~B}}{d x}\right)=\mathrm{R}\)
Using (3) and (4): ⇒ \(u_1 \frac{d \mathrm{~A}}{d x}+v_1 \frac{d \mathrm{~B}}{d x}=\mathrm{R}\)……..(8)
Solving (6) and (8) ⇒ \(\frac{d \mathrm{~A} / d x}{v \mathrm{R}}=\frac{d \mathrm{~B} / d x}{-u \mathrm{R}}=\frac{1}{v u_1-u v_1}\)
⇒ \(\frac{d \mathrm{~A}}{d x}=\frac{-v \mathrm{R}}{u v_1-v u_1}\) and \(\frac{d \mathrm{~B}}{d x}=\frac{u \mathrm{R}}{u v_1-v u_1}\)
A = \(\int \frac{-v \mathrm{R}}{u v_1-v u_1} d x\) and \(\mathrm{B}=\int \frac{u \mathrm{R}}{u v_1-v u_1} d x\)…..(9)
After integration, the constant is not added since A and B are involved in \(y_p\).
Substituting the values of A and B from (9) in (5), we get \(y_p\)
∴ The general solution of (1) is \(y=y_c+y_p \Rightarrow y=c_1 u+c_2 v+\mathrm{A} u+\mathrm{B} v\)
Another method: General solution of \(a_2 \frac{d^2 y}{d x^2}+a_1 \frac{d y}{d x}+a_0 y=Q(x)\) by the method of variation of parameters.
Given linear differential equation is \(a_2 \frac{d^2 y}{d x^2}+a_1 \frac{d y}{d x}+a_0 y=\mathrm{Q}(x)\)……(1)
Where \(a_2(\neq 0), a_1, a_0\) are functions of x or real constants and Q(x) is only a function of x. For the sake of convenience let \(a_2, a_1, a_0\) be real constants only.
The homogeneous equation corresponding to (1) is \(a_2 \frac{d^2 y}{d x^2}+a_1 \frac{d y}{d x}+a_0 y=0\)……(2)
Let \(y=y_c=c_1 y_1+c_2 y_2\) be the general solution of (2) where \(y_1\) and \(y_2\) are two L.I solutions of (2) and \(c_1, c_2\) are real constants. Hence it is the C.F. of (1).
Let P.I. of (1) be \(y_p=u y_1+v y_2 \ldots.\). (3) which is obtained from C.F. of (1) by replacing \(\dot{c}_1\) and \(c_2\) by u and v respectively which are also some functions of x and whose values are to be determined.
Differentiating the equation (3) twice, we get: \(y_p^{\prime}=\left(u y_1^{\prime}+v y_2^{\prime}\right)+\left(u^{\prime} y_1+v^{\prime} y_2\right) \ldots \ldots . . \text { (4) }\)
⇒ \(y_p^{\prime \prime}=\left(u y_1^{\prime \prime}+v y_2^{\prime \prime}\right)+\left(u^{\prime} y_1^{\prime}+v^{\prime} y_2^{\prime}\right)+\left(u^{\prime} y_1+v^{\prime} y_2\right)^{\prime} .\)……(5)
As \(y_p\) is a solution of (1), we have \(a_2 y_p^{\prime \prime}+a_1 y_p^{\prime \prime}+a_0 y_p=\mathrm{Q}(x)\)……..(6)
Substituting, (3), (4) and (5) in (6)
⇒ \(a_2\left[\left(u y_1^{\prime \prime}+v y_2^{\prime \prime}\right)+\left(u^{\prime} y_1^{\prime}+v^{\prime} y_2^{\prime}\right)+\left(u^{\prime} y_1+v^{\prime} y_2\right)^{\prime}\right]\)
+ \(a_1\left[\left(u y_1^{\prime}+v y_2^{\prime}\right)+\left(u^{\prime} y_1+v^{\prime} y_2\right)\right]+a_0\left(u y_1+v y_2\right)=\mathrm{Q}(x)\)
⇒ \(u\left(a_2 y_1^{\prime \prime}+a_1 y_1^{\prime}+a_0 y_1\right)+v\left(a_2 y_2^{\prime \prime}+a_1 y_2^{\prime}+a_0 y_2\right)\)
+ \(a_2\left(u^{\prime} y_1^{\prime}+v^{\prime} y_2^{\prime}\right)+a_2\left(u^{\prime} y_1+v^{\prime} y_2\right)^{\prime}+a_0 (u^{\prime} y_1+v^{\prime}\) \(y_2^{\prime})=\mathrm{Q}(x)\)…….(7)
Since \(y_1\) and \(y_2\) are the solutions of (2), we have : \(a_2 y_1^{\prime \prime}+a_1 y_1^{\prime}+a_0 y_1=0\) and \(a_2 y_2^{\prime \prime}+a_1 y_2^{\prime}+a_0 y_2=0\)…….(8)
(7) and (8) ⇒ \(a_2\left(u^{\prime} y_1^{\prime}+v^{\prime} y_2^{\prime}\right)+a_2\left(u^{\prime} y_1+v^{\prime} y_2\right)^{\prime}+a_1\left(u^{\prime} y_1+v^{\prime} y_2\right)=\mathrm{Q}(x)\)……..(9)
Choose u and v such that \(u^{\prime} y_1+v^{\prime} y_2=0\) …….(10)
(9) and (10) ⇒ \(a_2\left(u^{\prime} y_1^{\prime}+v^{\prime} y_2^{\prime}\right)=\mathrm{Q}(x) \Rightarrow u^{\prime} y_1^{\prime}+v^{\prime} y_2^{\prime}=\mathrm{Q}(x) / a_2\)……..(11)
Solving (10) and (11), we get :
⇒ \(u^{\prime}=-\left[y_2 \cdot \mathrm{Q}(x) / a_2\right] /\left(y_1 y_2^{\prime}-y_1^{\prime} y_2\right), v^{\prime}=\left[y_1 \cdot \mathrm{Q}(x) / a_2\right] /\left(y_1 y_2^{\prime}-y_1^{\prime} y_2\right)\)
⇒ \(u^{\prime}=\left|\begin{array}{cc}
0 & y_2 \\
\mathrm{Q}(x) / a_2 & y_2^{\prime}
\end{array}\right| /\left|\begin{array}{ll}
y_1 & y_2 \\
y_1^{\prime} & y_2^{\prime}
\end{array}\right|\), \(v^{\prime}=\left|\begin{array}{cc}
y_1 & 0 \\
y_1^{\prime} & \mathrm{Q}(x) / a_2
\end{array}\right| /\left|\begin{array}{ll}
y_1 & y_2 \\
y_1^{\prime} & y_2^{\prime}
\end{array}\right|\) ………………………. (12)
Integrating the equations in (12), we can find u and v. Substituting these u and v in (3), we get \(y_p\) of (1). Hence the general solution of (1) is \(y=y_c+y_p\).
Examples Of Non-Constant Coefficient Higher-Order Differential Equations
Example. Solve \(y^{\prime \prime}+2 y^{\prime}+y=e^{-x} \log x\)
Solution:
Given
\(y^{\prime \prime}+2 y^{\prime}+y=e^{-x} \log x\)G.E. in operator form is \(\left(D^2+2 D+1\right) y=e^{-x} \log x\)
A.E. of (1) is f(m)=0 \(\Rightarrow m^2+2 m+1=0 \Rightarrow(m+1)^2=0 \Rightarrow m=-1,-1\)
∴ \(y_c=\left(c_1+c_2 x\right) e^{-x}=c_1 e^{-x}+c_2 x e^{-x} \text {. }\)
Let \(y_1=e^{-x}, y_2=x e^{-x} \Rightarrow y_1^{\prime}=-e^{-x}, y_2^{\prime}=-x e^{-x}+e^{-x}\).
Let \(y_p=u y_1+v y_2\) where u=u(x), v=v(x) to be determined.
Now \(\left|\begin{array}{ll}y_1 & y_2 \\ y_1^{\prime} & y_2^{\prime}\end{array}\right|\)
= \(\left|\begin{array}{cc}e^{-x} & x e^{-x} \\ -e^{-x} & -x e^{-x}+e^{-x}\end{array}\right|=-x e^{-2 x}+e^{-2 x}+x e^{-2 x}=e^{-2 x}\)
Then \(u^{\prime}=\left|\begin{array}{cc}0 & x e^{-x} \\ -e^{-x} \cdot \log x & -x e^{-x}+e^{-x}\end{array}\right| / e^{-2 x}=\frac{x e^{-2 x} \log x}{e^{-2 x}}=-x \log x\)
⇒ \(v^{\prime}=\left|\begin{array}{cc}
e^{-x} & 0 \\
-e^{-x} & e^{-x} \log x
\end{array}\right| / e^{-2 x}\)
= \(\frac{e^{-2 x} \log x}{e^{-2 x}}=\log x\)
u = \(\int-x \log x d x=-\frac{x^2}{2} \log x+\frac{x^2}{4}, v=\int \log x d x=x \log x-x\)
∴ \(y_p=u y_1+v y_2=-\frac{x^2}{2} e^{-x} \log x+\frac{x^2}{4} e^{-x}(x \log x-x) x e^{-x}\)
Hence G.S. is \(y=y_c+y_p=\left(c_1+c_2 x\right) e^{-x}-\frac{x^2}{2} e^{-x} \log x+\frac{x^2}{4} e^{-x}+x^2 e^{-x} \log x-x^2 e^{-x}\)
⇒ \(y=\left(c_1+c_2 x\right) e^{-x}+\frac{1}{2} x^2 e^{-x} \log x-\frac{3}{4} x^2 e^{-x}\).
Note 1. The form of \(y_c\) and \(y_p\) is the same. But the constants that occur in $\(y_c\) are changed into functions of the independent variable x in \(y_p\). For this reason, the method of finding the P.I. is called the method of variation of parameters.
2. The above method can be extended to linear equations of order higher than the two.
3. The above method is applicable to linear equations with constant coefficients and also variable coefficients.
4. We know that the given linear equation of second order can be solved when an integral of C. F. is known.
Therefore the above method is surely superior to the variation of parameters since this method requires a complete knowledge of the C.F. instead of one integral of it.
Hence the method of variation of parameters should be used only when specifically asked to solve by this method.
Higher Order Linear Differential Equations 3 (Non-Constant Coefficients) Working Rule to find the general solution of \(\frac{d^2 y}{d x^2}+\mathrm{P} \frac{d y}{d x}+\mathrm{Q} y=\mathrm{R}\)… (1) by the method of variation of parameters :
1. In case the given equation is not in the standard form reduce it to the standard form.
2. Find the solution of \(\frac{d^2 y}{d x^2}+\mathrm{P} \frac{d y}{d x}+\mathrm{Q} y=0\).
Let its solution be \(y=c_1 u(x)+c_2 v(x)\) which is C. F of (1).
3. Let the P.I. of (1) be \(y_p=\mathrm{A} u+\mathrm{B} v\) where A and B are functions of x.
4. Find \(u \frac{d v}{d x}-v \frac{d u}{d x} \Rightarrow u v_1-v u_1\)
5. Find A and B by using: \(\mathrm{A}=\int \frac{-v \mathrm{R} d x}{u v_1-v u_1}, \mathrm{~B}=\int \frac{u \mathrm{R} d x}{u v_1-v u_1}\)
6. The general solution of (1) is \(y=c_1 u+c_2 v+\mathrm{A} u+\mathrm{B} v\)
Higher Order Linear Differential Equations III (Non-Constant Coefficients) Important: In Order To Solve A Linear Equation Of Second Order \(\frac{d^2 y}{d x^2}+\mathrm{P}(x) \frac{d y}{d x}+\mathrm{Q}(x) y=\mathrm{R}(x)\), Proceed As Mentioned Below
1. If the part of C. F. is given or known by inspection then apply the method.
2. If the method is mentioned in the given problem, then only apply the method of variation of parameters.
Higher Order Linear Differential Equations 3 (NonConstant Coefficients) Solved Problems
Example 1(a). Solve \(\left(\mathrm{D}^2+a^2\right) y=\tan a x\) by the method of variation of parameters.
Solution.
Given equation is \(\left(\mathrm{D}^2+a^2\right) y=\tan a x\)….(1)
where \(f(\mathrm{D}) \equiv \mathrm{D}^2+a^2\).
The A.E. is f(m)=0 \(\Rightarrow m^2+a^2=0 \Rightarrow m= \pm a i\).
∴ \(y_c=c_1 \cos a x+c_2 \sin a x \text {. }\)
Let the P.I. of (1) be \(y_p=A \cos a x+B \sin a x\) where A and B are functions of x……(2)
Then \(u=\cos a x, v=\sin a x, \mathrm{R}=\tan a x\).
By the method of variation of parameters: \(u v_1-v u_1=\cos a x(a \cos a x)-\sin a x(-a \sin a x)=a\left(\cos ^2 a x+\sin ^2 a x\right)=a\)
Now \(\mathrm{A}=\int \frac{-v \mathrm{R}}{u v_1-v u_1} d x=-\int \frac{\sin a x \tan a x}{a} d x=-\frac{1}{a} \int \frac{\sin ^2 a x}{\cos a x} d x\)
= \(-\frac{1}{a} \int \frac{1-\cos ^2 a x}{\cos a x} d x=-\frac{1}{a}\left[\int \sec a x d x-\int \cos a x d x\right]\)
= \(-\frac{1}{a^2} \log |\sec a x+\tan a x|+\frac{1}{a^2} \sin a x\)
B = \(\int \frac{u \mathrm{R}}{u v_1-v u_1} d x=\int \frac{\cos a x \tan a x}{a} d x=\frac{1}{a} \int \sin a x d x=-\frac{1}{a^2} \cos a x\)
(2), (3), (4) ⇒ \(y_p=\frac{1}{a^2}[\sin a x-\log |\sec a x+\tan a x|] \cos a x\)
–\(\frac{1}{a^2} \cos a x \cdot \sin a x=-\frac{1}{a^2} \log |\sec a x+\tan a x| \cdot \cos a x\)
∴ The general solution of (1) is ⇒ \(y=y_c+y_p \Rightarrow y=c_1 \cos a x+c_2 \sin a x-\frac{1}{a^2} \cos a x \log (\sec a x+\tan a x)\)
Example 1(b). Solve \(\left(D^2+4^2\right) y=\tan 2 x\) by the method of variation of parameters.
Solution: Put a=2 in the above example.
Methods For Solving Non-Constant Coefficient Linear Differential Equations
Example 2. Solve \(\left(D^2+1\right) y=\ cosec x\) by the method of variation of parameters.
Solution:
Given equation is \(\left(\mathrm{D}^2+1\right) y=\text{cosec} x\) where f(D) \(\equiv \mathrm{D}^2+1\)
The A.E. is f(m)=0 \(\Rightarrow m^2+1=0 \Rightarrow m= \pm i\)
∴ \(y_c=c_1 \cos x+c_2 \sin x\)
Let the P.I. of (1) be \(y_p=\text{Acos} x+\mathrm{B} \sin x\)
where A = \(\mathrm{A}(x), \mathrm{B}=\mathrm{B}(x), u=\cos x, v=\sin x\) and \(\mathrm{R}=\text{cosec} x\)
Now by the method of variation of parameters: \(u v_1-v u_1=(\cos x)(\cos x)-(\sin x)(-\sin x)=\cos ^2 x+\sin ^2 x=1\)
A = \(\int \frac{-v \mathrm{R}}{u v_1-v u_1} d x=-\int \frac{\sin x \text{cosec} x}{1} d x=-\int d x=-x\)
B = \(\int \frac{u \mathrm{R}}{u v_1-v u_1} d x=\int \frac{\cos x \text{cosec} x}{1} d x=\int \cot x d x=\log |\sin x|\)
(2), (3), (4) ⇒ \(y_p=(-x) \cos x+(\log |\sin x|) \sin x\)
∴ The general solution of (1) is \(y=y_c+y_p\)
⇒ \(y=c_1 \cos x+c_2 \sin x-x \cos x+\sin x \log |\sin x|\)
Example 3. Solve \(\left(\mathrm{D}^2-2 \mathrm{D}\right) y=e^x \sin x\) by the method of variation of parameters.
Solution:
Given equation is \(\left(\mathrm{D}^2-2 \mathrm{D}\right) y=e^x \sin x\)…..(1)
where \(f(\mathrm{D}) \equiv \mathrm{D}^2-2 \mathrm{D}\)
The A.E. is \(f(m)=0 \Rightarrow m^2-2 m=0 \Rightarrow m(m-2)=0 \Rightarrow m=0,2\)
∴ \(y_c=c_1+c_2 e^{2 x}\)
Let the P.I. of (1) be \(y_p=\mathrm{A}+\mathrm{B} e^{2 x}\)…….(2)
where A = \(\mathrm{A}(x), \mathrm{B}=\mathrm{B}(x), u=1, v=e^{2 x}\) and \(\mathrm{R}=e^x \sin x\)
Now by the method of variation of parameters: \(u v_1-v u_1=1\left(2 e^{2 x}\right)-e^{2 x}(0)=2 e^{2 x}\)
A = \(\int \frac{-v \mathrm{R}}{u v_1-v u_1} d x=-\int \frac{e^{2 x} \cdot e^x \sin x}{2 e^{2 x}} d x=-\frac{1}{2} \int e^x \sin x d x\)
= \(-\frac{1}{2}\left[\frac{e^x \cdot \sin x-e^x \cos x}{1^2+1^2}\right]=\frac{1}{4} e^x(\cos x-\sin x) \ldots \ldots \ldots \ldots(3)\)…….(3)
B = \(\int \frac{u \mathrm{R}}{u v_1-v u_1} d x=\int \frac{1\left(e^x \sin x\right)}{2 e^{2 x}} d x=\frac{1}{2} \int e^{-x} \sin x d x\)
= \(\frac{1}{2}\left[\frac{e^{-x}(-1) \sin x-e^{-x} \cos x}{(-1)^2+1^2}\right]=-\frac{1}{4} e^{-x}(\cos x+\sin x) \ldots \ldots \ldots\)(4)
(2), (3), (4) ⇒ \(y_p=\frac{1}{4} e^x(\cos x-\sin x)-\frac{1}{4} e^{-x}(\cos x+\sin x) e^{2 x}\)
⇒ \(y_p=\frac{1}{4} e^x(\cos x-\sin x-\cos x-\sin x)=-\frac{1}{2} e^x \sin x\)
∴ The general solution of (1) is \(y=y_c+y_p=c_1+c_2 e^{2 x}-\frac{1}{2} e^x \sin x\)
Higher Order Differential Equations Non-Constant Coefficients Solved Problems
Example 4. Solve \(\left[(x-1) \mathrm{D}^2-x \mathrm{D}+1\right] y=(x-1)^2\) by the method of variation of parameters.
Solution:
Given equation in the standard form is \(\frac{d^2 y}{d x^2}-\frac{x}{x-1} \frac{d y}{d x}+\frac{1}{x-1} y=x-1\)…….(1)
Homogeneous equation of (1) is \(\frac{d^2 y}{d x^2}-\frac{x}{x-1} \frac{d y}{d x}+\frac{1}{x-1} y=0\) ……….(2)
where P = \(\frac{-x}{x-1}, \mathrm{Q}=\frac{1}{x-1}\) and \(\mathrm{R}=(x-1)\) from (1) \(1+\mathrm{P}+\mathrm{Q}=1-\frac{x}{x-1}+\frac{1}{x-1}=0 \Rightarrow y=e^x\) is a solution of (2).
Also \(\mathrm{P}+\mathrm{Q} x=-\frac{x}{x-1}+\frac{x}{x-1}=0 \Rightarrow y=x\) is a solution of (2)
∴ \(y_c \text { of }(1)=c_1 e^x+c_2 x\)
Let the P.I. of (1) be \(y_p=\mathrm{A} e^x+\mathrm{B} x\)……..(3)
where A = \(\mathrm{A}(x), \mathrm{B}=\mathrm{B}(x), u=e^x, v=x\) and R=x-1
Now by the method of variation of parameters: \(u v_1-v u_1=e^x(1)-x\left(e^x\right)=(1-x) e^x\)
A = \(\int \frac{-v \mathrm{R}}{u v_1-v u_1} d x=\int \frac{-x(x-1)}{(1-x) e^x} d x=\int x e^{-x} d x=-x e^{-x}+\int e^{-x} d x\)
= \(-x e^{-x}-e^{-x}=-(1+x) e^{-x} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\)…….(4)
B = \(\int \frac{u \mathrm{R}}{u v_1-v u_1} d x=\int \frac{e^x(x-1)}{e^x(1-x)} d x=-\int d x=-x\)…..(5)
(3), (4), (5) ⇒ \(y_p=-(1+x) e^{-x} \cdot e^x-x(x)=-\left(1+x+x^2\right)\)
∴ The general solution of (1) is \(y=y_c+y_p \Rightarrow y=c_1 e^x+c_2 x-\left(1+x+x^2\right)\)
Example 5. Solve \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y=x^2 e^x\) by the method of variation of parameters.
Solution:
Given equation in the standard form is \(\frac{d^2 y}{d x^2}+\frac{1}{x} \frac{d y}{d x}-\frac{1}{x^2} y=e^x\)…….(1)
and its homogeneous equation is \(\frac{d^2 y}{d x^2}+\frac{1}{x} \frac{d y}{d x}-\frac{1}{x^2} y=0\)……(2)
where P = \(\frac{1}{x}, \mathrm{Q}=-\frac{1}{x^2}\) and \(\mathrm{P}+\mathrm{Q} x=\frac{1}{x}-\frac{1}{x}=0 \Rightarrow x\) is a part of C.F.
Put y = v x where v=v(x) \(\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} ; \frac{d^2 y}{d x^2}=x \frac{d^2 v}{d x^2}+2 \frac{d v}{d x}\)……(3)
Now (2) and (3) ⇒ \(x \frac{d^2 v}{d x^2}+2 \frac{d v}{d x}+\frac{v}{x}+\frac{d v}{d x}-\frac{1}{x^2}(v x)=0 \Rightarrow x \frac{d^2 v}{d x^2}+3 \frac{d v}{d x}=0\)
⇒ \(\int \frac{d^2 v}{d x^2}+\frac{3}{x} \frac{d v}{d x}=0 \Rightarrow \int \frac{\left(d^2 v / d x^2\right)}{(d v / d x)}=-\int \frac{3}{x} d x \Rightarrow \log \frac{d v}{d x}=-3 \log x+\log c\)
⇒ \(\log \frac{d v}{d x}=\log \frac{c}{x^3} \Rightarrow \frac{d v}{d x}=\frac{c}{x^3} \Rightarrow \int \frac{d v}{d x}=c \int \frac{1}{x^3} d x+c_1 \Rightarrow v=-\frac{c}{2 x^2}+c_1\)
C.F. of the equation (1) is \(y=v x=c_1 x-\frac{c_2}{x} \Rightarrow y_c=\) C.F. of \((1)=c_1 x+\left(c_2 / x\right)\)
Alter to find the part of C. F.
Put \(x=e^z\) or \(z=\log x\). Let \(\frac{d}{d z}=\theta \Rightarrow[\theta(\theta-1)+\theta-1] y=0 \Rightarrow \theta^2-1=0 \Rightarrow \theta= \pm 1\)
Then \(y_c=c_1 e^z+c_2 e^{-z}=c_1 x+\left(c_2 / x\right)\)
Let the P.I. of (1) be \(y_p=\mathrm{A} x+(\mathrm{B} / x)\)………(4)
where A = \(\mathrm{A}(x), \mathrm{B}=\mathrm{B}(x), u=x, v=1 / x\) and R = \(e^x\) [from (1)]
Now by the method of variation of parameters: \(u v_1-v u_1=x\left(-1 / x^2\right)-(1 / x) 1=-(1 / x)-1 / x=-2 / x\)
A = \(\int \frac{-v \mathrm{R}}{u v_1-v u_1} d x=\int \frac{-(1 / x) e^x}{(-2 / x)} d x=\frac{1}{2} \int e^x d x=\frac{1}{2} e^x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \)(5)
B = \(\int \frac{u \mathrm{R}}{u v_1-v u_1} d x=\int \frac{x \cdot e^x}{(-2 / x)} d x=-\frac{1}{2} \int x^2 e^x d x=-\frac{1}{2} x^2 e^x+x e^x-e^x\)
(4), (5), (6) ⇒ \(y_p=\frac{1}{2} x e^x-\frac{1}{2} x e^x+e^x-\frac{1}{x} e^x=e^x-\frac{1}{x} e^x=e^x\left(\frac{x-1}{x}\right)\)
∴ The G.S. of (1) is \(y=y_c+y_p=c_1 x+\left(c_2 / x\right)+e^x(x-1) / x\)
Applications Of Non-Constant Coefficient Linear Differential Equations
Example 6. solve \(\left(\mathrm{D}^2-3 \mathrm{D}+2\right) y=\cos \left(e^{-x}\right)\)
Solution:
Given equation is \(\left(\mathrm{D}^2-3 \mathrm{D}+2\right) y=\cos \left(e^{-x}\right)\)……(1)
where \(f(\mathrm{D}) \equiv \mathrm{D}^2-3 \mathrm{D}+2\)
The A:E. is f(m)=0 ⇒ \(m^2-3 m+2=0\) (m-1)(m-2)=0, m=1,2
∴ \(y_c=c_1 e^x+c_2 e^{2 x}\)
Let the P.I. of (1) be \(y_p=A e^x+B e^{2 x}\)…….(2)
where A = \(\mathrm{A}(x), \mathrm{B}=\mathrm{B}(x), u=e^x, y=e^{2 x}\) and R = \(\cos \left(e^{-x}\right)\) (from (1)) Now by the method of variation of parameters : \(u v_1-v u_1=e^x\left(2 e^{2 x}\right)-e^{2 x}\left(e^x\right)=2 e^{3 x}-e^{3 x}=e^{3 x}\)
A = \(\int \frac{-v \mathrm{R}}{u_1-v u_1} d x=\int \frac{-e^{2 x} \cos \left(e^{-x}\right)}{e^{3 x}} d x=-\int e^{-x} \cos \left(e^{-x}\right) d x\)
= \(\int \cos t d t=\sin t=\sin \left(e^{-x}\right)\left[\mathrm{Put} e^{-x}=t \Rightarrow-e^{-x} d x=d t\right] \ldots \ldots \ldots(3)\)
B = \(\int \frac{u \mathrm{R}}{u v_1-\dot{v} u_1} d x=\int \frac{e^x \cos \left(e^{-x}\right)}{e^{3 x}} d x=\int e^{-2 x} \cos e^{-x} d x\)
= \(-\int t \cos t d t=-\left[t(\sin t)-\int \sin t d t\right]\left[\text { Put } e^{-x}=t \Rightarrow-e^{-x} d x=d t\right]\)
= \(-t \sin t+(-\cos t)=-\left(e^{-x} \sin e^{-x}+\cos e^{-x}\right) \ldots \ldots \ldots \ldots(4)\)…….
(2), (3), (4) \(\Rightarrow y_p=e^x \sin e^{-x}-\left(e^{-x} \sin e^{-x}+\cos e^{-x}\right) e^{2 x}\)
= \(e^x \sin e^{-x}-e^x \sin e^{-x}-e^{2 x} \cos e^{-x}=-e^{2 x} \cos e^{-x}\)
∴ The G.S.of (1) is \(y=y_c+y_p=c_1 e^x+c_2 e^{2 x}-e^{2 x} \cos \left(e^{-x}\right)\)
xample 7. If y=x and \(y=x e^{2 x}\) are L.I. solutions of the homogeneous equation corresponding to \(x^2 \frac{d^2 y}{d x^2}-2 x(1+x) \frac{d y}{d x}+2(x+1) y=x^3\), solve it by the method of variation of parameters.
Solution:
Given equation in the standard form is \(\frac{d^2 y}{d x^2}-\frac{2(1+x)}{x} \frac{d y}{d x}+\frac{2(x+1)}{x^2} y=x\)…….(1)
and its homogeneous equation is \(\frac{d^2 y}{d x^2}-\frac{2(1+x)}{x} \frac{d y}{d x}+\frac{2(x+1)}{x^2} y=0\)…….(2)
Given y=x and \(y=x e^{2 x}\) are L.I. solutions of (2)
∴ \(y_c=\) C.F. of (1)=\(c_1 x+c_2 x e^{2 x}\)
Let the P.I. of (1) be \(y_p=\mathrm{A} x+\mathrm{B} x e^{2 x}\)…….(3)
where A = \(\mathrm{A}(x), \mathrm{B}=\mathrm{B}(x), u=x, v=x e^{2 x}\) and R=x (from (1))
Now by the method of variation of parameters: \(u v_1-v u_1=x\left(e^{2 x}+2 x e^{2 x}\right)-x e^{2 x}(1)=2 x^2 e^{2 x}\)
A = \(\int \frac{-v \mathrm{R}}{u v_1-v u_1} d x=-\int \frac{x e^{2 x} \cdot x}{2 x^2 e^{2 x}} d x=-\frac{1}{2} \int d x=-\frac{x}{2} \ldots \ldots(4)\)…….(4)
B = \(\int \frac{u \mathrm{R}}{u v_1-v u_1} d x=\int \frac{x(x)}{2 x^2 e^{2 x}} d x=\frac{1}{2} \int e^{-2 x} d x=-\frac{1}{4} e^{-2 x} \ldots \ldots \ldots \ldots .\)…….(5)
(3), (4), (5) \(y_p=(-x / 2) x+(-1 / 4) e^{-2 x} \cdot x e^{2 x}=-(1 / 2) x^2-(1 / 4) x\)
∴ The G.S. of (1) is \(y=y_c+y_p=c_1 x+c_2 x e^{2 x}-(1 / 2) x^2-(1 / 4) x\)
Example 8. solve \(\left(D^2-2 D+2\right) y=e^x \tan x\) by the method of variation of parameters.
Solution:
Given equation is \(\left(\mathrm{D}^2-2 \mathrm{D}+2\right) y=e^x \tan x\) where \(f(\mathrm{D}) \equiv \mathrm{D}^2-2 \mathrm{D}+2\)………(1)
The A.E. is f(m)=0 \(\Rightarrow m^2-2 m+2=0^{\circ}\)
m = \(\frac{2 \pm \sqrt{4-8}}{2}=\frac{2 \pm 2 i}{2}=1-i, 1+i\) are the roots of A . E
∴ \(y_c=\) C.F. of (1) \(=e^x\left(c_1 \cos x+c_2 \sin x\right)\)
Let the P.I. of (1) be \(y_p=\mathrm{Ae} e^x \cos x+\mathrm{Be} e^x \sin x\)………(2)
where \(\mathrm{A}=\mathrm{A}(x), \mathrm{B}=\mathrm{B}(x), u=e^x \cos x, v=e^x \sin x\) and \(\mathrm{R}=e^x \tan x\)
Now by the method of variation of parameters: \(u v_1-v u_1=e^x \cos x\left(e^x \cos x+e^x \sin x\right)-e^x \sin x\left(e^x \cos x-e^x \sin x\right)\)
= \(e^{2 x} \cos ^2 x+e^{2 x} \sin x \cos x-e^{2 x} \sin x \cos x+e^{2 x} \sin ^2 x=e^{2 x}\)
A = \(\int \frac{-v \mathrm{R}}{u v_1-v u_1} d x=-\int \frac{e^x \sin x \cdot e^x \cdot \tan x}{e^{2 x}} d x=-\int \frac{\sin ^2 x}{\cos x}=-\int \frac{1-\cos ^2 x}{\cos x} d x\)
= \(\int(\cos x-\sec x) d x=\sin x-\log |\sec x+\tan x| \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\)………(3)
B = \(\int \frac{u \mathrm{R} d x}{u v_1-v u_1}=\int \frac{e^x \cos x \cdot e^x \tan x}{e^{2 x}} d x=\int \sin x d x=-\cos x \ldots \ldots \ldots \ldots\)……..(4)
∴ The G.S. of (1) is y = \(y_c+y_p=e^x\left(c_1 \cos x+c_2 \sin x\right)-e^x \cos x \log |\sec x+\tan x|\)
Properties Of Higher-Order Differential Equations With Non-Constant Coefficients
Example 9. solve \(y^{\prime \prime}+3 y^{\prime}+2 y=12 e^x\) by the method of variation of par ameters.
Solution:
Given \(\left(\mathrm{D}^2+3 \mathrm{D}+2\right) y=12 e^x\)……(1)
A.E. is \(\mathrm{D}^2+3 \mathrm{D}+2=0 \Rightarrow(\mathrm{D}+2)(\mathrm{D}+1)=0 \Rightarrow \mathrm{D}=-2,-1\)
∴ \(y_c=c_1 e^{-2 \dot{x}}+c_2 e^{-x}\).
Let P.I. of (1) be \(y_p=\mathrm{A} e^{-2 x}+\mathrm{Be} e^{-x}\).
Where A = \(\mathrm{A}(x), \mathrm{B}=\mathrm{B}(x), u=e^{-2 x}, \mathrm{~V}=e^{-x}\) and \(\mathrm{R}=12 e^x\).
Now by the method of variation of parameters : \(u \mathrm{~V}_1-\mathrm{V} u_1=e^{-2 x}\left(-e^{-x}\right)-e^{-x}\left(-2 e^{-2 x}\right)=e^{-3 x}\)
A = \(\int \frac{-\mathrm{VR} d x}{u \mathrm{~V}_1-\mathrm{V} u_1}=\int \frac{-e^{-x} \cdot 12 e^x}{e^{-3 x}} d x=-12 \int e^{3 x} d x=-\frac{12 e^{3 x}}{3}=-4 e^{3 x}\)
B = \(\int \frac{u \mathrm{R} d x}{u \mathrm{~V}_1-\mathrm{V} u_1}=\int \frac{e^{-2 x} \cdot 12 e^x d x}{e^{-3 x}}=12 \int e^{2 x} d x=\frac{12 e^{2 x}}{2}=6 e^{2 x}\)
∴ \(y_p=-4 e^{3 x} \cdot e^{-2 x}+6 e^{2 x} \cdot e^{-x}=6 e^x-4 e^x=2 e^x\)
∴ The G.S. of (1) is \(y=c_1 e^{-2 x}+c_2 e^{-x}+2 e^x\)
Worked Examples Of Higher-Order Differential Equations With Variable Coefficients
Example 10. solve \(y^{\prime \prime}-2 y^{\prime}+y=e^x \log x\) by the method of variation of parameters.
Solution:
Given \(y^{\prime \prime}-2 y^{\prime}+y=e^x \log x \Rightarrow\left(D^2-2 D+1\right) y=e^x \log x\)…….(1)
A.E. is \(D^2-2 D+1=0 \Rightarrow(D-1)^2=0 \Rightarrow D=1,1\)
∴ \(y_c=\left(c_1+c_2 x\right) e^x\)
Let P.I. of (1) be \(\cdot y_p=\mathrm{A} e^x+\mathrm{B} x e^x\) where \(u=e^x, v=x e^x \mathrm{R}=e^x \log x\)
Now \(u v_1-v u_1=e^x \cdot\left(e^x+x e^x\right)-x e^x \cdot e^x=e^{2 x}+x e^{2 x}-x e^{2 x}=e^{2 x}\)
A = \(\int \frac{\mathrm{VR}}{\mathrm{uv}_1-v u_1} d x=\int \frac{x e^x \cdot e^x \log x}{e^{2 x}} d x=\int x \log x d x\)
= \(\frac{x^2}{2} \log x-\int \frac{x^2}{2} \cdot \frac{1}{x} d x=\frac{x^2}{2} \log x-\frac{1}{2} \frac{x^2}{2}=\frac{x^2}{2} \log x-\frac{x^2}{4}\)
B = \(\int \frac{u \mathrm{R} d x}{u v_1-v u_1}=\int \frac{e^x \cdot e^x \log x}{e^{2 x}} d x=\int \log x d x=x \log x-x\)
∴ \(y_p=\left(\frac{x^2}{2} \log x-\frac{x^2}{4}\right) e^x+(x \log x-x) x e^x\)
G.S: of (1) is \(y=y_c+y_p=\left(c_1+c_2 x\right) e^x+\left(\frac{x^2}{2} \log x-\frac{x^2}{4}\right) e^x+(x \log x-x) x e^x\)
Higher Order Linear Differential Equations III (Non-Constant Coefficients) Exercise 6(a)
Solve the following differential equation by the method of variation of parameters:
1. \(\left(\mathrm{D}^2+a^2\right) y=\cos a x\)
Solution: \(y=c_1 \cos a x+c_2 \sin a x+(x / 2 a) \sin a x+\left(1 / 4 a^2\right) \cos a x\)
2. \(\left(D^2+1\right) y=\sec x\)
Solution: \(y=c_1 \cos x+c_2 \sin x+(\cos x) \log \cos x+x \sin x\)
3. \(\left(\mathrm{D}^2+a^2\right) y=\sec a x\)
Solution: \(y=c_1 \cos a x+c_2 \sin a x+\left(1 / a^2\right)(\cos a x) \log \cos x+(1 / a) x \sin a x\)
4. \(y^{\prime \prime}+4 y=4 \sec ^2 2 x\)
Solution: \(y=c_1 \cos 2 x+c_2 \sin 2 x-1+(\sin 2 x) \log |\sec 2 x+\tan 2 x|\)
5. \(\frac{d^2 y}{d x^2}+4 y=4 \tan 2 x\)
Solution: \(y=c_1 \cos 2 x+c_2 \sin 2 x-(\cos 2 x) \log |\sec 2 x+\tan 2 x|\)
6. \(\left(D^2+1\right) y=\ cosec x \cot x\)
Solution: \(y=c_1 \cos x+c_2 \sin x-(\cos x) \log \sin x-(\cot x+x) \sin x\)
7. \(\left(\mathrm{D}^2+1\right) y=x \cos x\)
Solution: \(y=c_1 \cos x+c_2 \sin x+\frac{x}{4} \cos x+\frac{x^2}{4} \sin x\)
8. \(\left(\mathrm{D}^2+a^2\right) y=\ cosec a x\)
Solution: \(y=c_1 \cos a x+c_2 \sin a x-(1 / a) x \cos a x+\left(1 / a^2\right)(\sin a x) \log \sin a x\)
9. \(\frac{d^2 y}{d x^2}-y=\frac{2}{1+e^x}\)
Solution: \(y=c_1 e^x+c_2 e^{-x}-1+e^x \log \left(e^{-x}+1\right)-e^{-x} \log \left(e^x+1\right)\)
10. \(\left(D^2-1\right) y=\left(1+e^{-x}\right)^{-2}\)
Solution: \(y=c_1 e^x+c_2 e^{-x}-1+e^{-x} \log \left(1+e^x\right)\)
11. \(\left(\mathrm{D}^2-1\right) y=e^{-x} \sin \left(e^{-x}\right)+\cos \left(e^{-x}\right)\)
Solution: \(y=c_1 e^x+c_2 e^{-x}-e^x \sin e^{-x}\)
12. \(y^{\prime \prime}+2 y^t+y=x^2 e^{-x}\)
Solution: \(y=c_1 e^{-x}+c_2 x e^{-x}+(1 / 12) x^4 e^{-x}\)
13. \(\left(\mathrm{D}^2-3 \mathrm{D}+2\right) y=\sin \left(e^{-x}\right)\)
Solution: \(y=c_1 e^x+c_2 e^{2 x}-e^{2 x} \sin \left(e^{-x}\right)\)
14. \((1-x) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y=2(x-1)^2 e^{-x}, 0<x<1\) given that y=x and \(y=e^x\) are L.I. solutions of the homogeneous equation corresponding to the given equation.
Solution: \(y=c_1 x+c_2 e^x+[(1 / 2)-x] e^{-x}\)
Higher Order Linear Differential Equations III (Non Constant Coefficients) Cauchy-Euler Equation
An equation of the form \(x^n \frac{d^n y}{d x^n}+\mathrm{P}_1 x^{n-1} \frac{d^{n-1} y}{d x^{n-1}}+\cdots \ldots+\mathrm{P}_n y=\mathrm{Q}\)
Where, \(P_1 \ldots . P_n\) are real constants and Q is a function of x defined on an interval I is called a homogenous linear equation or Cauchy Euler equation of order n and its operator form is \(v\left(x^n D^n+\mathrm{P}_1 D^{n-1}+\cdots \ldots+\mathrm{P}_n\right) y=\mathrm{Q}(x)\) where \(\mathrm{D} \equiv \frac{d}{d x}\)
Cauchy Euler equation can be transformed into a linear equation with constant coefficients by the change of independent variable with the substitution \(x=e^z\)
or \(z=\log x, x>0\)
⇒ \(\frac{d z}{d x}=\frac{1}{x} \text { Now } \frac{d y}{d x}=\frac{d y}{d z} \frac{d z}{d x}=\frac{1}{x} \frac{d y}{d z} \Rightarrow x \frac{d y}{d x}=\frac{d y}{d z}\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{1}{x} \frac{d y}{d z}\right)=-\frac{1}{x^2} \frac{d y}{d z}+\frac{1}{x} \frac{d}{d x}\left(\frac{d y}{d z}\right)=-\frac{1}{x^2} \frac{d y}{d z}+\frac{1}{x} \frac{d}{d z}\left(\frac{d y}{d z}\right) \frac{d z}{d x}\)
= \(-\frac{1}{x^2} \frac{d y}{d z}+\frac{1}{x^2} \frac{d y}{d z} \Rightarrow x^2 \frac{d^2 y}{d x^2}=\frac{d^2 y}{d z^2}-\frac{d y}{d z}\)
⇒ \(\frac{d^3 y}{d x^3}=\frac{d}{d x}\left(-\frac{1}{x^2} \frac{d y}{d z}+\frac{1}{x^2} \frac{d^2 y}{d z^2}\right)=\frac{2}{x^3} \frac{d y}{d z}-\frac{1}{x^2} \frac{d^2 y}{d z^2} \frac{1}{x}-\frac{2}{x^3} \frac{d^2 y}{d z^2}+\frac{1}{x^2} \frac{d^3 y}{d z^3} \frac{1}{x}\)
= \(\frac{1}{x^3} \frac{d^3 y}{d z^3}-\frac{3}{x^3} \frac{d^2 y}{d z^2}+\frac{2}{x^3} \frac{d y}{d z}=x^3 \frac{d^3 y}{d x^3}=\frac{d^3 y}{d z^3}-3 \frac{d^2 y}{d z^2}+2 \frac{d y}{d z}\)
We have \(\frac{d}{d x} \equiv \mathrm{D}\).
Let the differential operator \(\frac{d}{d z}\) be denoted by \(\theta\) so that \(\frac{d}{d z} \equiv \theta\).
Then \(\frac{d^2}{d z^2} \equiv \theta^2, \frac{d^3}{d z^3} \equiv \theta^3, \ldots \ldots, \frac{d^n}{d z^n} \equiv \theta^n\)
(1) (2), (3) ⇒ \(x \frac{d y}{d x}=x \mathrm{D} y=\theta y, x^2 \frac{d^2 y}{d x^2}=x^2 \mathrm{D}^2 y=\theta^2 y-\theta y=\theta(\theta-1) y\)
⇒ \(x^3 \frac{d^3 y}{d x^3}=x^3 \mathrm{D}^3 y=\theta^3 y-3 \theta^2+2 \theta y=\theta(\theta-1)(\theta-2) y\)
⇒ \(x^{n-1} \frac{d^{n-1}}{d x^{n-1}}=x^{n-1} \mathrm{D}^{n-1} y=[\theta(\theta-1) \ldots \ldots\{\theta-(n-2)\}] y\)
⇒ \(x^n \frac{d^n}{d x^n}=x^n \mathrm{D}^n y=[0(\theta-1) \ldots \ldots\{\theta-(n-1)\}] y\)
(5) \(\mathrm{D} \equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1), x^3 \mathrm{D}^3 \equiv \theta(\theta-1)(\theta-2), \ldots . x^n D^n \equiv \theta(\theta-1)\) …….(n-1)
Substituting the values from equation (5) in Cauchy Euler equation, we get \(\theta(\theta-1) \ldots \ldots(\theta-n+1) y+P_1 \theta(\theta-1) \ldots \ldots(\theta-n+2) y+\cdots .+P_n y=Q\left(e^z\right)\)
⇒ \(\left[\theta(\theta-1) \ldots(\theta-n+1)+P_1 \theta(\theta-1) \ldots(\theta-n+2)+\cdots+P_n\right] y=Q\left(e^z\right) \Rightarrow f(\theta) y=z\)
where \(f(\theta) \equiv \theta(\theta-1) \ldots(\theta-n+1)+P_1 \theta(\theta-1) \ldots(\theta-n+2)+\cdots .+P_n\) and \(z=Q\left(e^z\right)\)
This is a linear differential equation with constant coefficients.
The differential operator \(f(\theta)\) and the inverse operator \(\frac{1}{f(\theta)}\) obey the properties of f (D) and \(\frac{1}{f(\mathrm{D})}\).
Hence \(f(\theta) y=z\). can be solved by the methods discussed already in this chapter.
Higher Order Linear Differential Equations III (Non-Constant Coefficients) Solved problems
Example 1: Solve \(3 x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=x\)
Solution:
Given Equation in operator form is \(\left(3 x^2 \mathrm{D}^2+x \mathrm{D}+1\right) y=x\)…….(1)
which is a homogenous linear equation. Put \(x=e^z \Rightarrow z=\log x,^{\prime} x>0\).
Let \(\theta \equiv \frac{d}{d z}\) then \(x \mathrm{D} \equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1)\)…….(2)
(1) and (2) \(\Rightarrow[3 \theta(\theta-1)+\theta+1] y=e^z \Rightarrow\left(3 \theta^2-2 \theta+1\right) y=e^z\)…..(3)
where \(f(\theta) \equiv 3 \theta^2-2 \theta+1\). Then A.E. is \(f(m)=0 \Rightarrow 3 m^2-2 m+1=0\)…….(4)
⇒ \(m=\frac{2 \pm \sqrt{4-12}}{6}=\frac{2 \pm 2 \sqrt{2} i}{6}=\frac{1}{3} \pm \frac{i \sqrt{2}}{3}\) are the roots of (4)
∴ \(y_c=e^{z / 3}\left(c_1 \cos \frac{\sqrt{2}}{3} z+c_2 \sin \frac{\sqrt{2}}{3} z\right)\)
⇒ \(y_p=\frac{1}{3 \theta^2-2 \theta+1} e^z=e^z \frac{1}{3\left(1^2\right)-2(1)+1}=\frac{1}{2} e^z\)
The G.S. of (3) is \(y=y_c+y_p \Rightarrow y=e^{z / 3}\left(c_1 \cos \frac{\sqrt{2}}{3} z+c_2 \sin \frac{\sqrt{2}}{3} z\right)+\frac{e^z}{2}\)
∴ The general solution of (1) is \(y=x^{1 / 3}\left[c_1 \cos \left(\frac{\sqrt{2} \log x}{3}\right)+c_2 \sin \left(\frac{\sqrt{2} \log x}{3}\right)\right]+\frac{x}{2}\)
(because \(e^{z / 3}=e^{(1 / 3) \log x}=e^{\log x^{1 / 3}}=x^{1 / 3}\)and \(e^z=e^{\log x}=x\))
Example 2: Solve \(x^3 \frac{d^3 y}{d x^3}+2 x^2 \frac{d^2 y}{d x^2}+2 y=10\left(x+\frac{1}{x}\right)\)
Solution:
Given Equation in operator form is \(\left(x^3 \mathrm{D}^3+2 x^2 \mathrm{D}^2+2\right) y=10\left(x+\frac{1}{x}\right)\)……(1)
Let \(x=e^z \Rightarrow z=\log x, x>0\) and \(\theta \equiv \frac{d}{d z}\).
Then x D \(\equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1), x^3 \mathrm{D}^3 \equiv \theta(\theta-1)(\theta-2)\)
(1) and (2) \(\Rightarrow[\theta(\theta-1)(\theta-2)+2 \theta(\theta-1)+2] y=10\left(e^z+e^{-z}\right)\)
⇒ \(\left(\theta^3-\theta^2+2\right) y=10\left(e^z+e^{-z}\right)\) where \(f(\theta) \equiv \theta^3-\theta^2+2\)……(3)
The A.E. is \(f(m)=0 \Rightarrow m^3-m^2+2=0 \Rightarrow(m+1)\left(m^2-2 m+2\right)=0\)
⇒ \(m+1=0, m^2-2 m+2=0 \Rightarrow m=-1, m=\frac{2 \pm \sqrt{4-8}}{2}=\frac{2 \pm 2 i}{2}=1 \pm i\)
-1,1 \(\pm i\) are the roots of (3). ∴ \(y_c=c_1 e^{-z}+e^z\left(c_2 \cos z+c_3 \sin z\right)\)
⇒ \(y_p=10 \frac{1}{\theta^3-\theta^2+2}\left(e^z+e^{-z}\right)=10\left[\frac{1}{\theta^3-\theta^2+2} e^z+\frac{1}{\theta^3-\theta^2+2} e^{-z}\right]\)
= \(10 \frac{e^z}{1-1+2}+10 \frac{1}{(\theta+1)\left(\theta^2-2 \theta+2\right)} e^{-z}=5 e^z+\frac{10}{1+2+2} \cdot \frac{1}{\theta+1} e^{-z}=5 e^z+2 . z e^{-z}\)
The G.S. of (3) is \(y=y_c+y_p \Rightarrow y=c_1 e^{-z}+e^z\left(c_2 \cos z+c_3 \sin z\right)+5 e^z+2 z e^{-z}\)
∴ The G.S. of (1) is \(y=c_1 x^{-1}+x\left[c_2 \cos (\log x)+c_3 \sin (\log x)\right]+5 x+\left(\frac{2}{x}\right) \log x\)
Example 3: Solve \(\left(x^2 \mathrm{D}^2+2 x \mathrm{D}-12\right) y=x^3 \log x\)
Solution:
Given Equation \(\left(x^2 \mathrm{D}^2+2 x \mathrm{D}-12\right) y=x^3 \log x\)……(1)
Let \(x=e^z \Rightarrow z=\log x, x>0\) and \(\theta \equiv \frac{d}{d z} \Rightarrow x \mathrm{D} \equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1)\)…….(2)
(1) and (2) \(\Rightarrow[\theta(\theta-1)+2 \theta-12] y=z e^{3 z} \Rightarrow\left(\theta^2+\theta-12\right) y=z e^{3 z}\)
where \(f(\theta) \equiv \theta^2+\theta-12\)……..(3)
The A.E. is f(m)=0 \(\Rightarrow m^2+m-12=0\)
(m-3)(m+4)=0 ⇒ m=3,-4 are the roots of (3).
∴ \(y_c=c_1 e^{3 z}+c_2 e^{-4 z}\)
⇒ \(y_p=\frac{1}{(\theta-3)(\theta+4)} z e^{3 z}=e^{3 z} \frac{1}{(\theta+3-3)(\theta+3+4)} z=e^{3 z} \frac{1}{\theta(\theta+7)} z\)
= \(e^{3 z} \frac{1}{\theta+7}\left(\frac{1}{\theta} z\right)=e^{3 z} \frac{1}{\theta+7} \frac{z^2}{2}=\frac{e^{3 z}}{14}\left(1+\frac{\theta}{7}\right)^{-1} z^2\)
= \(\frac{e^{3 z}}{14}\left(1-\frac{\theta}{7}+\frac{\theta^2}{49}\right) z^2=\frac{e^{3 z}}{14}\left(z^2-\frac{2 z}{7}+\frac{2}{49}\right)=\frac{e^{3 z}}{98}\left(7 z^2-2 z\right)+\frac{1}{343} e^{3 z}\)
G.S. of (2) is \(y=y_c+y_p=c_1 e^{3 z}+c_2 e^{-4 z}+\frac{e^{3 z}}{98}\left(7 z^2-2 z\right)\)
(Note: \((1 / 343) e^{3 z}\) is abosrbed in \(c_1 e^{3 z}\))
∴ G.S. of (1) is \(y=c_1 x^3+c_2 \frac{1}{x^4}+\frac{x^3}{98}\left[7(\log x)^2-2(\log x)\right]\)
Example 4: Solve \(\left(x^4 \mathrm{D}^3+2 x^3 \mathrm{D}^2-x^2 \mathrm{D}+x\right) y=1\)
Solution:
Given Equation is \(\left(x^4 \mathrm{D}^3+2 x^3 \mathrm{D}^2-x^2 \mathrm{D}+x\right) y=1\)…..(1)
Dividing by \(x(>0)\) ⇒ \(\left(x^3 \mathrm{D}^3+2 x^2 \mathrm{D}^2-x \mathrm{D}+1\right) y=\frac{1}{x}\) which is clearly homogeneous …….(1)
Let \(x=e^z \Rightarrow z^{\prime}=\log x, x>0\) and \(\theta \equiv \frac{d}{d z}\)
⇒ \(x \mathrm{D} \equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1), x^3 \mathrm{D}^3 \equiv \theta(\theta-1)(\theta-2)\)……(3)
(1) and (2) ⇒ \([\theta(\theta-1)(\theta-2)+2 \theta(\theta-1)-\theta+1] y=e^{-z}\)
⇒ \((\theta-1)\left(\theta^2-1\right) y=e^{-z}\) where \(f(\theta) \equiv(\theta-1)\left(\theta^2-1\right)\)
The A.E. is f(m)=0 \(\Rightarrow(m-1)\left(m^2-1\right)=0 \Rightarrow m=1,1,-1\) are the roots.
∴ \(y_c=\left(c_1+c_2 z\right) e^z+c_3 e^{-z}\)
∴ \(y_p=\frac{1}{(\theta-1)^2(\theta+1)} e^{-z}=\frac{1}{(-1-1)^2} \cdot \frac{1}{\theta+1} e^{-z}=\frac{1}{4} z e^{-z}\)
The G. S. of (3) is \(y=y_c+y_p \Rightarrow y=\left(c_1+c_2 z\right) e^z+c_3 e^{-z}+(1 / 4) z e^{-z}\)
∴ The G. S. of (1) is \(y=\left(c_1+c_2 \log x\right) x+c_3(1 / x)+(\log x) /(4 x)\)
Example 5: Solve \(x^2 \frac{d^2 y}{d x^2}-3 x \frac{d y}{d x}+5 y=x^2 \sin (\log x)\)
Solution:
Given Equation in the operator form is \(\left(x^2 \mathrm{D}^2-3 x \mathrm{D}+5\right) y=x^2 \sin (\log x)\)…….(1)
Let \(x=e^z \Rightarrow z=\log x, x>0\) and \(\frac{d}{d z} \equiv \theta \Rightarrow x \mathrm{D} \equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1)\)……..(2)
(1) and (2) \(\Rightarrow[\theta(\theta-1)-3 \theta+5] y=e^{2 z} \sin z \Rightarrow\left(\theta^2-4 \theta+5\right) y=e^{2 z} \sin z\)…..(3)
where \(f(\theta) \equiv \theta^2-4 \theta+5\)
The A.E. is f(m)=0 \(\Rightarrow m^2-4 m+5=0\)……(3)
m=\(\frac{4 \pm \sqrt{16-20}}{2}=\frac{4 \pm 2 i}{2}=2 \pm i\)
∴ \(y_c=e^{2 z}\left(c_1 \cos z+c_2 \sin z\right)\)
⇒ \(y_p=\frac{1}{\theta^2-4 \theta+5}\left(e^{2 z} \sin z\right)=e^{2 z} \frac{1}{(\theta+2)^2-4(\theta+2)+5} \sin z\)
= \(e^{2 z} \cdot \frac{1}{\theta^2+1} \sin z=e^{2 z}\left(\frac{-z}{2}\right) \cos z\)
∴ The G. S. of (3) is \(y=y_c+y_p \Rightarrow y=e^{2 z}\left(c_1 \cos z+c_2 \sin z\right)-\frac{z}{2} e^{2 z} \cos z\)
∴ The G. S. of (1) is \(y=x^2\left[c_1 \cos (\log x)+c_2 \sin (\log x)\right]-\frac{1}{2}(\log x) \cdot x^2 \cos (\log x)\)
Example 6: Solve \(\left(x^2 \mathrm{D}^2+x \mathrm{D}+1\right) y=(\log x) \sin (\log x)\)
Solution:
Given Equation \(\left(x^2 \mathrm{D}^2+x \mathrm{D}+1\right) y=(\log x) \sin (\log x)\)……..(1)
Let \(x=e^z \Rightarrow z=\log x, x>0\) and \(\frac{d}{d z} \equiv \theta \Rightarrow x \mathrm{D} \equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1)\)…….(2)
(1) and (2) \(\Rightarrow[\theta(\theta-1)+\theta+1] y=z \sin z \Rightarrow\left(\theta^2+1\right) y=z \sin z\)……(3)
where \(f(\theta) \equiv \theta^2+1\).
The A.E. is f(m)=0 \\(Rightarrow m^2+1=0 \Rightarrow m= \pm i\)……..(4)
⇒ \(m=-i, i\) are the roots of (4)
∴ \(y_c=c_1 \cos z+c_2 \sin z\)
⇒ \(y_p=\frac{1}{\theta^2+1} z \sin z=\) I.P. of \(\frac{1}{\theta^2+1} z e^{i z}=\) I.P. of \(e^{i z} \frac{1}{(\theta+i)^2+1} z\)
= I.P. of \(e^{i z} \frac{1}{\theta^2+2 i \theta} z=\) I.P. of \(e^{i z} \cdot \frac{1}{2 i \theta}\left(1+\frac{\theta}{2 i}\right)^{-1} z\)
= I.P. of \(e^{i z} \frac{1}{2 i \theta}\left(1-\frac{i \theta}{2}\right)^{-1} z=\) I.P. of \(e^{i z} \cdot \frac{1}{2 i \theta}\left[1+\frac{i \theta}{2}-\frac{\theta^2}{4}+\cdots\right] z\)
= I.P. of \(e^{i z}\left(\frac{-i}{2}\right)\left(\frac{1}{\theta}+\frac{i}{2}-\frac{\theta}{4}+\cdots ..\right) z=\) I.P. of \(e^{i z}\left(\frac{-i}{2}\right)\left(\frac{z^2}{2}+\frac{i z}{2}-\frac{1}{4}\right)\)
= \(-\frac{1}{2}\) I.P. of \((\cos z+i \sin z)\left(\frac{z^2 i}{2}-\frac{z}{2}-\frac{i}{4}\right)\)
= \(-\frac{1}{2}\left[\left(\frac{z^2}{2}-\frac{1}{4}\right) \cos z-\frac{z}{2} \sin z\right]=-\frac{z^2}{4} \cos z+\frac{z}{4} \sin z+\frac{1}{8} \cos z\)
∴ The G. S. of (3) is \(y=c_1 \cos z+c_2 \sin z-\frac{z^2}{4} \cos z+\frac{z}{4} \sin z\)
(because(1 / 8) cos z is absorbed in \(y_c\))
∴ The G. S. of (1) is \(y=c_1 \cos (\log x)+c_2 \sin (\log x)-\frac{(\log x)^2}{4} \cos (\log x)+\frac{\log x}{4} \sin (\log x)\)
Example 7: Solve \(x^2 \frac{d^2 y}{d x^2}+3 x \frac{d y}{d x}+y=\frac{1}{(1-x)^2}\)
Solution:
Given Equation in operator form is \(\left(x^2 \mathrm{D}^2+3 x \mathrm{D}+1\right) y=\frac{1}{(1-x)^2}\)…….(1)
Let \(x=e^z \Rightarrow z=\log x, x>0\) and \(\frac{d}{d z} \equiv \theta \Rightarrow x \mathrm{D} \equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1)\)……(2)
(1) and (2) \([\theta(\theta-1)+3 \theta+1] y=\frac{1}{\left(1-e^z\right)^2} \Rightarrow\left(\theta^2+2 \theta+1\right) y=\frac{1}{\left(1-e^z\right)^2}\)…..(3)
where \(f(\theta) \equiv \theta^2+2 \theta+1\)……4)(
The A.E. is f(m)=0 \(\Rightarrow m^2+2 m+1=0\)
⇒ \((m+1)^2=0 \Rightarrow m=-1,-1\) are the roots of (4)
∴ \(y_c=\left(c_1+c_2 z\right) e^{-z}\)
⇒ \(y_p=\frac{1}{(\theta+1)^2\left(1-e^z\right)^2}=\frac{1}{(\theta+1)}\left[\frac{1}{(\theta+1)} \cdot \frac{1}{\left(1-e^z\right)^2}\right]=\frac{1}{(\theta+1)}\left[e^{-z} \int \frac{1}{\left(1-e^z\right)^2} \cdot e^z d z\right]\)
(\(\left[\frac{1}{\mathrm{D}+\alpha} \mathrm{Q}=e^{-\alpha z} \int \mathrm{Q} e^{\alpha z} d z\right]=\frac{1}{\theta+1}\left[e^{-z} \cdot \int \frac{d t}{(1-t)^2}\right]\)) where \(t=e^z \Rightarrow d t=e^z d z\)
= \(\frac{1}{\theta+1} e^{-z} \cdot \frac{1}{1-t}=\frac{1}{\theta+1} \frac{e^{-z}}{1-e^z}=e^{-z} \int \frac{e^{-z}}{1-e^z} \cdot e^z d z\)(using the same formula)
= \(e^{-z} \int \frac{d z}{1-e^z}=e^{-z} \int \frac{e^{-z}}{e^{-z}-1} d z=-e^{-z} \log \left|e^{-z}-1\right|\)
∴ The G.S. of (3) is \(y=y_c+y_p \Rightarrow y=\left(c_1+c_2 z\right) e^{-z}-e^{-z} \log \left|e^{-z}-1\right|\)
∴ The G. S. of (1) is \(y=\left(c_1+c_2 \log x\right) \frac{1}{x}-\frac{1}{x} \log \left|\frac{1-x}{x}\right|\).
Example 8: Solve \(x^2 \frac{d^2 y}{d x^2}+4 x \frac{d y}{d x}+2 y=e^x\)
Solution:
Given Equation in operator form is \(\left(x^2 \mathrm{D}^2+4 x \mathrm{D}+2\right) y=e^x\)…..(1)
Let \(x=e^z \Rightarrow z=\log x, x>0\) and \(\frac{d}{d z} \equiv \theta \Rightarrow x \mathrm{D} \equiv \theta, x^2 \mathrm{D}^2 \equiv \theta(\theta-1)\)…..(2)
(1) and (2) \(\Rightarrow[\theta(\theta-1)+4 \theta+2] y=\exp \left(e^z\right) \Rightarrow\left(\theta^2+3 \theta+2\right) y=\exp \left(e^z\right)\)…….(3)
where \(f(\theta) \equiv \theta^2+3 \theta+2\)
The A.E. is f(m)=0 \(\Rightarrow m^2+3 m+2=0\)…….(4)
⇒ \((m+2)(m+1)=0 \Rightarrow m=-1,-2\) are the roots of (4)
∴ \(y_c=c_1 e^{-z}+c_2 e^{-2 z}\)
⇒ \(y_p=\frac{1}{(\theta+1)(\theta+2)} \exp \left(e^z\right)=\left(\frac{1}{\theta+1}-\frac{1}{\theta+2}\right) \exp \left(e^z\right)\)
Let \(\frac{1}{\theta+1} \exp \left(e^z\right)=u \Rightarrow(\theta+1) u=\exp \left(e^z\right) \Rightarrow \frac{d u}{d z}+u=\exp \left(e^z\right)\) which is a linear in u.
I.F. = \(e^z[/latex
∴ [latex]u e^z=\int e^z \cdot e^{e^z} d z=e^{e^z}\)
(because \(e^z=x, e^z d z=d x\) and \(\int e^x d x=e^x=e^{e^z}\))
u = \(e^{-z} \cdot \exp \left(e^z\right)\)
Let \(\frac{1}{\theta+2} \exp \left(e^z\right)=v \Rightarrow(\theta+2) v=\exp \left(e^z\right) \Rightarrow \frac{d v}{d z}+2 v=\exp \left(e^z\right)\)
I.F. = \(e^{2 z}\)
∴ \(v e^{2 z}=\int e^{2 z} \cdot e^{e^z} \cdot d z=\int x e^x d x\) (because \(e^z=x \Rightarrow e^z d x=d x\) and \(\exp \left(e^z\right)=e^x\))
⇒ \(v e^{2 z}=e^x(x-1)=\left(e^z-1\right) e^{e^z} \Rightarrow v=e^{-2 z}\left(e^z-1\right) e^{e^z}\)
∴ \(y_p=e^{-z} \cdot e^{e^z}-e^{-2 z}\left(e^z-1\right) e^{e^z}=e^{e^z}\left(e^{-z}-e^{-z}+e^{-2 z}\right)=e^{e^z} \cdot e^{-2 z}\)
Now G. S. of (3) is \(y=c_1 e^{-z}+c_2 e^{-2 z}+e^{e^z} \cdot e^{-2 z}\)
∴ G.S. of (1) is \(y=\frac{c_1}{x}+\frac{c_2}{x^2}+\frac{e^x}{x^2}\)
Aliter to find \(y_p\)
⇒ \(y_p=\frac{1}{(\theta+1)(\theta+2)} \exp \left(e^z\right)=\frac{1}{(\theta+2)}\left[\frac{1}{\theta+1} e^{e^z}\right]=\frac{1}{\theta+2}\left[e^{-z} \int e^{e^z} \cdot e^z d z\right] \)
= \(\frac{1}{\theta+2}\left[e^{-z} \cdot e^{e^z}\right]=e^{-2 z} \cdot \int e^{-z} \cdot e^{u^z} \cdot e^{2 z} d z=e^{-2 z} \int e^{e^z} \cdot e^z d z=e^{-2 z} \cdot e^{e^z}=e^x / x^2\)
Higher Order Linear Differential Equations III (Non Constant Coefficients) Exercise 6(b)
1. \(\left(x^2 \mathrm{D}^2-x \mathrm{D}+1\right) y=\log x\)
Solution: \(y=\left(c_1+c_2 \log x\right) x+\log x+2\)
2. (a) \(x^2 \frac{d^2 y}{d x^2}-3 x \frac{d y}{d x}+4 y=2 x^2\)
Solution: \(y=x^2\left[c_1+c_2 \log x+(\log x)^2\right]\)
(b). \(\left(x^2 \mathrm{D}^2+x \mathrm{D}-4\right) y=x^2\)
Solution: \(y=c_1 x^2+c_2 x^{-2}+(1 / 4) x^2 \log x\)
3. \(\left(x^2 \mathrm{D}^2-x \mathrm{D}+1\right) y=2 \log x\)
Solution: \(y=x\left(c_1+c_2 \log x\right)+2 \log x+4\)
4. \(\left(x^2 \mathrm{D}^2+x \mathrm{D}-1\right) y=x^3\)
Solution: \(y=c_1 x+\left(c_2 / x\right)+\left(x^3 / 8\right)\)
5. \(\left(x^2 \mathrm{D}^2+2 x \mathrm{D}-20\right) y=(x+1)^2\)
Solution: \(y=c_1 x^{-5}+c_2 x^4-(1 / 14) x^2-(1 / 9) x-(1 / 20)\)
6. \(\left(x^2 \mathrm{D}^2+2 x \mathrm{D}-20\right) y=x+\log x\)
Solution: \(y=c_1 x^{-5}+c_2 x^4-(1 / 8) x-(1 / 400)(20 \log x-1)\)
7. \(\left(x^2 \mathrm{D}^2-2 x \mathrm{D}+2\right) y=x+\left(\frac{1}{x}\right)\)
Solution: \(y=c_1 x+c_2 x^2-x \log x+(1 / 6 x)\)
8. \(\frac{d^2 y}{d x^2}+\frac{1}{x} \frac{d y}{d x}=\frac{12 \log x}{x^2}\)
Solution: \(y=c_1+c_2 \log x+2(\log x)^3\)
9. \(\left(x^2 \mathrm{D}^3+3 x \mathrm{D}^2+\mathrm{D}\right) y=x^2 \log x\)
Solution: \(y=c_1+c_2(\log x)+c_3(\log x)^2+\left(x^3 / 27\right)(\log x-1)\)
10. \(\left(x^2 \mathrm{D}^2+x \mathrm{D}+4\right) y=2 x l_n x \quad(x>0)\)
Solution: \(y=c_1 \cos (2 \log x)+c_2 \sin (2 \log x)+(2 / 5) x[\log x-(2 / 5)]\)
11. \(\left(x^2 \mathrm{D}^2-x \mathrm{D}-3\right) y=x^2 \log x\)
Solution: \(y=c_1 x^{-1}+c_2 x^3-\left(x^2 / 9\right)(3 \log x+2)\)
12. \(\left(x^3 \mathrm{D}^3+2 x^2 \mathrm{D}^2+x \mathrm{D}-1\right) y=\cos (\log x)\)
Solution: \(y=c_1 x+c_2 \cos (\log x)+c_3 \sin (\log x)-(1 / 4)(\log x) \cos (\log x)\)\(-(1 / 4)(\log x) \cdot \sin (\log x)\)
13. \(\left(x^3 \mathrm{D}^3-x^2 \mathrm{D}^2+2 x \mathrm{D}-2\right) y=x^3+3 x\)
Solution: \(y=\left(c_1+c_2 \log x\right) x+c_3 x^2+(1 / 4) x^3-(3 / 2) x(\log x)^2\)
14. \(\left(x^3 \mathrm{D}^3+3 x^2 \mathrm{D}^2+x \mathrm{D}+8\right) y=65 \cos (\log x)\)
Solution: \(y=c_1 x^{-2}+\left[c_2 \cos (\sqrt{3} \log x)+c_3 \sin (\sqrt{3} \log x)\right] x\)\(+8 \cos (\log x)-\sin (\log x)\)
15. \(\left(x^2 \mathrm{D}^2-x \mathrm{D}+4\right) y=\cos (\log x)+x \sin (\log x)\)
Solution: \(y=x\left[c_1 \cos (\sqrt{3} \log x)+c_2 \sin (\sqrt{3} \log x)\right]+\frac{1}{13}[3 \cos (\log x)\) \(-2 \sin (\log x)]+\left(\frac{x}{2}\right) \sin (\log x)\)
16. \(\left(x^3 \mathrm{D}^3+x^2 \mathrm{D}^2\right) y=1+x+x^2\)
Solution: \(y=c_1+\left(c_2+c_3 \log x\right) x+\log x+(x / 2)(\log x)^2+\left(x^2 / 2\right)\)
17. \(\left(x^2 \mathrm{D}^2-x \mathrm{D}+2\right) y=x \log x\)
Solution: \(y=x\left[c_1 \cos (\log x)+c_2 \sin (\log x)\right]+x \log x\)
18. \(\left(x^3 \mathrm{D}^3+2 x^2 \mathrm{D}^2+3 x \mathrm{D}-3\right) y=x^2+x\)
Solution: \(y=c_1 x+c_2 \cos (\sqrt{3} \log x)+c_3 \sin (\sqrt{3} \log x)+\left(x^2 / 7\right)+(x / 4) \log x\)
19. \(\left(x^2 \mathrm{D}^2+x \mathrm{D}-1\right) y=\frac{1}{1+x}\)
Solution: \(y=c_1 x+c_2 x^{-1}+(x / 2) \log \left(1+x^{-1}\right)-(1 / 2 x) \log (1+x)-(1 / 2)\)
20.(a) \(\left(x^3 \mathrm{D}^3+3 x^2 \mathrm{D}^2+x \mathrm{D}+1\right) y=x \log x\)
Solution: \(y=\frac{c_1}{x}+\sqrt{x}\left[c_2 \cos \left(\frac{\sqrt{3}}{2}\right) \log x+c_3 \sin \left(\frac{\sqrt{3}}{2} \log x\right)\right]+\frac{x}{2} \log x-\frac{3 x}{4}\)
(b). \(x^3 \mathrm{D}^3 y+3 x^2 \mathrm{D}^2 y+x \mathrm{D} y+y=x+\log x\)
Solution: \(y=c_1 x^{-1}+\sqrt{x}\left[c_2 \cos \left(\frac{\sqrt{3}}{2} \log x\right)+c_3 \sin \left(\frac{\sqrt{3}}{2} \log x\right)\right]+\log x+\frac{1}{2} x\)
Higher Order Linear Differential Equations III (Non Constant Coefficients) Legender’s Equation
An equation of the form \((a x+b)^n \frac{d^n y}{d x^n}+P_1(a x+b)^{n-1} \frac{d^{n-1} y}{d x^{n-1}}+\cdots .+P_n y=Q\) where \(P_1, P_2, \ldots ., P_n\) are real constants and Q is a function of x defined on an interval I is called Legender’s linear equation.
Such equations can be reduced to linear equations with constant coefficients by the substitution \(a x+b=e^z \Rightarrow z=\log (a x+b) \Rightarrow \frac{d z}{d x}=\frac{a}{a^2+b} \text {. }\)
Now \(\frac{d y}{d x}=\frac{d y}{d z} \cdot \frac{d z}{d x}=\frac{a}{a x+b} \frac{d y}{d z} \Rightarrow(a x+b) \frac{d y}{d x}=a \frac{d y}{d z}=a \theta y\)
where \(\theta \equiv \frac{d}{d z} \Rightarrow(a x+b) \mathrm{Dy}=a \theta y\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left[\frac{a}{a x+b} \frac{d y}{d z}\right]=\frac{d y}{d z} \cdot a \frac{d}{d x}\left(\frac{1}{a x+b}\right)+\frac{a}{a x+b} \cdot \frac{d}{d z}\left(\frac{d y}{d z}\right) \cdot \frac{d z}{d x}\)
= \(\frac{-a^2}{(a x+b)^2} \frac{d y}{d z}+\frac{a^2}{(a x+b)^2} \frac{d^2 y}{d z^2}=\frac{a^2}{(a x+b)^2}\left(\frac{d^2 y}{d z^2}-\frac{d y}{d z}\right)\)
⇒ \((a x+b)^2 \cdot \frac{d^2 y}{d x^2}=a^2\left(\theta^2 y-\theta y\right)=a^2 \theta(\theta-1) y \Rightarrow(a x+b)^2 D^2 y=a^2 \theta(\theta-1) y\)
Similarly, \((a x+b)^3 \frac{d^3 y}{d x^3}=a^3 \theta(\theta-1)(\theta-2) y\)
⇒ \((a x+b)^3 \mathrm{D}^3 y=a^3 \theta(\theta-1)(\theta-2) y\) and so on.
Higher Order Linear Differential Equations III (Non Constant Coefficients) Solved Problems
1. Solve \(\left[(1+x)^2 \mathrm{D}^2+(1+x) \mathrm{D}+1\right] y=4 \cos \log (1+x)\)
Solution:
Given equation is \((1+x)^2 \mathrm{D}^2 y+(1+x) \mathrm{D} y+y=4 \cos \log (1+x)\)……..(1)
Let \(1+x=e^z \Rightarrow z=\log (1+x)\)
and \(\frac{d}{d z} \equiv \theta \Rightarrow(x+1) \mathrm{D} \equiv \theta,(x+1)^2 \mathrm{D}^2 \equiv \theta(\theta-1) \text {. }\)…….(2)
(1) and (2) ⇒ \(\theta(\theta-1) y+\theta y+y=4 \cos z \Rightarrow\left(\theta^2+1\right) y=4 \cos z\)……(3)
where \(f(\theta) \equiv \theta^2+1\)
The A.E. is f(m)=0 \(\Rightarrow m^2+1=0 \Rightarrow m=-i, i\)
∴ \(y_c=c_1 \cos z+c_2 \sin z=c_1 \cos \log (1+x)+c_2 \sin \log (1+x)\).
⇒ \(y_p=4 \cdot \frac{1}{\theta^2+1} \cos z=4 \cdot \frac{z}{2} \sin z=2 z \sin z=2 \log (1+x) \cdot \sin \log (1+x)\).
∴ The G.S. of (3) is \(y=y_c+y_p \Rightarrow y=c_1 \cos z+c_2 \sin z+2 z \sin z\).
∴ The G.S. of (1) is y = \(c_1 \cos \log (1+x)+c_2 \sin \log (1+x)+2[\log (1+x)] \sin \log (1+x)\)
2. Solve \((3 x+2)^2 \frac{d^2 y}{d x^2}+3(3 x+2) \frac{d y}{d x}-36 y=3 x^2+4 x+1\)
Solution:
The given equation in the operator form is \(\left[(3 x+2)^2 D^2+3(3 x+2) \mathrm{D}-36\right] y=3 x^2+4 x+1\)…..(1)
Let \((3 x+2)=e^z \Rightarrow z=\log (3 x+2)\) and \(\frac{d}{d z}=\theta \Rightarrow x=\frac{e^z-2}{3}\)
Also \((3 x+2) \mathrm{D} \equiv 3 \theta\), and \((3 x+2)^2 \mathrm{D}^2=3^2 \theta(\theta-1)\)……(2)
(1) and (2) } \(\Rightarrow\left[3^2 \theta(\theta-1)+3.3 \theta-36\right] y=3\left(\frac{e^z-2}{3}\right)^2+4\left(\frac{e^z-2}{3}\right)+1\)
⇒ \(\left(9 \theta^2-9 \theta+9 \theta-36\right) y=\frac{1}{3}\left(e^{2 z}+4-4 e^z+4 e^z-8+3\right)\)
⇒ \(\left(\theta^2-4\right) y=\frac{1}{27}\left(e^{2 z}-1\right)\) where\(f(\theta)=\theta^2-4 \text {. }\)……..(3)
The A.E. is f(m)=0 \(\Rightarrow m^2-4=0 \Rightarrow m=-2,2\)
∴ \(y_c=c_1 e^{2 z}+c_2 e^{-2 z}=c_1(3 x+2)^2+c_2\left[1 /(3 x+2)^2\right]\)
⇒ \(y_p=\frac{1}{27} \frac{1}{\theta^2-4}\left(e^{2 z}-1\right)=\frac{1}{27}\left[\frac{1}{\theta^2-4} e^{2 z}-\frac{1}{\theta^2-4} e^{0 z}\right]\)
= \(\frac{1}{27}\left[\frac{1}{2+2} \cdot \frac{1}{\theta-2} e^{2 z}-\frac{1}{0-4}\right]=\frac{1}{27}\left[\frac{1}{4} \cdot z e^{2 z}+\frac{1}{4}\right]\)
= \(\frac{1}{108}\left(z e^{2 z}+1\right)=\frac{1}{108}\left[(3 x+2)^2 \log (3 x+2)+1\right]\)
G.S. of (3) is \(y=y_c+y_p=c_1 e^{2 z}+c_2 e^{-2 z}+\frac{1}{108}\left(z e^{2 z}+1\right)\)
G.S. of (1) is \(y=c_1(3 x+2)^2+c_2(3 x+2)^{-2}+(1 / 108)\left[(3 x+2)^2 \log (3 x+2)+1\right]\)
3. Solve \(\left[(1+2 x)^2 D^2-6(1+2 x) D+16\right] y=8(1+2 x)^2\)
Solution:
Given \(\left[(1+2 x)^2 D^2-6(1+2 x) D+16\right] y=8(1+2 x)^2\)…..(1)
Let \(1+2 x=e^z \Rightarrow z=\log (1+2 x)\) and \(\frac{d}{d z}=\theta \Rightarrow x=\frac{e^z-1}{2}\)
Also \((1+2 x) \mathrm{D} \equiv 2 \theta\) and \((1+2 x)^2 \mathrm{D}^2 \equiv 2^2 \theta(\theta-1)\)…..(2)
(1) and (2) \(\Rightarrow\left[2^2 \theta(\theta-1)-6.2 \theta+16\right] y=8 e^{2 z}\)
⇒ \(\left(4 \theta^2-16 \theta+16\right) y=8 e^{2 z} \Rightarrow\left(\theta^2-4 \theta+4\right)^y=2 e^{2 z}\)….(3)
where \(f(\theta)=\theta^2-4 \theta+4\). A.E. is f(m)=0 ⇒ \(m^2-4 m+4=0\)
⇒ \((m-2)^2=0 \Rightarrow m=2,2\)
∴ \(y_c=\left(c_1+c_2 z\right) e^{2 z}=\left[c_1+c_2 \log (1+2 x)\right](1+2 x)^2\)
⇒ \(y_p=2 \frac{1}{(\theta-2)^2} e^{2 z}=2 \cdot \frac{z^2}{2!} e^{2 z}=[\log (1+2 x)]^2(1+2 x)^2\)
The G.S. of (3) is \(y=y_c+y_p=\left(c_1+c_2 z\right) e^{2 z}+z^2 e^{2 z}\)
The G.S. of (1) is \(y=\left[c_1+c_2 \log (1+2 x)\right](1+2 x)^2+[\log (1+2 x)]^2(1+2 x)^2\).
Higher Order Linear Differential Equations III (Non Constant Coefficients) Exercise 6(c)
Solve the following differential equations.
1. \(\left[(5+2 x)^2 \mathrm{D}^2-6(5+2 x) \mathrm{D}+8 y=0\right.\)
Solution: \(y=(5+2 x)^2\left[c _ { 1 } \cos h \left(\sqrt{2} \log (5+2 x)+c_2 \sin h(\sqrt{2} \log (5+2 x)]\right.\right.\)
2. \((1+x)^2 \frac{d^2 y}{d x^2}+(1+x) \frac{d y}{d x}+y=2 \sin \log (1+x)\)
Solution: \(y=c_1 \cos \log (1+x)+c_2 \sin \log (1+x)-[\log (x+1)]\)\(\cos \log (x+1)\)
3. \((x+1)^2 \frac{d^2 y}{d x^2}-3(x+1) \frac{d y}{d x}+4 y=x^2+x+1\)
Solution: \(y=\left[c_1+c_2 \log (x+1)\right](x+1)^2+(1 / 2)[\log (x+1)]^2\)\((x+1)^2-(x+1)+(1 / 4)\)
4. \(\left[(x+3)^2 \mathrm{D}^2-4(x+3) \mathrm{D}+6\right] y=\log (x+3)\)
Solution: \(y=c_1(x+3)^2+c_2(x+3)^3+(1 / 36)[6 \log (x+3)+5]\)
5. \(\left[(x+a)^2 \mathrm{D}^2-4(x+a) \mathrm{D}+6\right] y=x .\)
Solution: \(y=c_1(x+a)^2+c_2(x+a)^3+\frac{x+a}{2}-\frac{a}{6}\)
6. \(\left[(2 x+1)^2 \mathrm{D}^2-2(2 x+1) \mathrm{D}-12\right] y=6 x\)
Solution: \(y=c_1(2 x+1)^3+c_2(2 x+1)^{-1}-(3 / 16)(2 x+1)+(1 / 4)\)
7. \(\left[(2 x+3)^2 \mathrm{D}^2-2(2 x+3) \mathrm{D}-12\right] y=6 x\)
Solution: \(y=c_1(2 x+3)^{-1}+c_2(2 x+3)^3-(3 / 16)(2 x+3)+(3 / 4)\)
Higher Order Linear Differential Equations III (Non-Constant Coefficients) Miscellaneous Differential Equations
Type 1: Differential equations of the form \(\frac{d^2 y}{d x^2}=f(x)\)
To solve such kinds of equations, integrating this equation once w.r.t. x on both sides we get \(\frac{d y}{d x}=\int f(x) d x+c_1\).
Here \(c_1\) is an integrating constant.
⇒ \(\frac{d y}{d x}=\mathrm{F}(x)+c_1 \text {. }\)
Here F(x)=\int f(x) d x
Again integrating this equation we get
y = \(\int F(x) d x+c_1 x+C\), where C is an integrating constant.
This becomes the general solution of the given differential equation consisting of two arbitrary constants.
If the given differential equation is of the form \(\frac{d^3 y}{d x^3}=F(x)\) then integrating this equation thrice successively we get the general solution consisting of 3 arbitrary constants.
In general, to solve the equation of the form \(\frac{d^n y}{d x^n}=f(x)\) we have to integrate it n times successively by which we can get a general solution of the given equation consisting of n arbitrary constants.
Type 2: Equations of the form \(\frac{d^2 y}{d x^2}=f(y)=\) function of y.
Such types of equations generally occur in dynamics.
Higher Order Linear Differential Equations III (Non-Constant Coefficients) Solved Problems
1. Solve \(\frac{d^2 y}{d x^2}=x^2 \sin x\)
Solution:
The given differential equation is \(\frac{d^2 y}{d x^2}=x^2 \sin x\)
Integrating with respect to x we get \(\frac{d y}{d x}=\int x^2 \sin x d x+c_1\).
Here \(c_1\) is an integrating constant.
⇒ \(\frac{d y}{d x}=\left(x^2\right) \int \sin x d x-\int 2 x \cdot\left(\int \sin x d x\right)+c_1\), using Integration by parts
⇒ \(\frac{d y}{d x}=-x^2 \cos x+2 \int x \cos x d x+c_1\)
⇒ \(\frac{d y}{d x}=-x^2 \cos x+2\left[x \sin x-\int \cos x d x\right]+c_1\)
⇒ \(\frac{d y}{d x}=-x^2 \cos x+2[x \sin x-\sin x d x]+c_1\)
⇒ \(\frac{d y}{d x}=-x^2 \cos x+2 x \sin x-2 \sin x d x+c_1\)
Integrating with respect to x as an y = \(\int\left[-x^2 \cos x-2(1-x) \sin x+c_1\right] d x+c\)
Here c is an integrating constant.
y = \(-\left[x^2 \sin x-\int 2 x \sin x d x\right]-2 \int(1-x) \sin x d x+c_1 x+c\)
y = \(-x^2 \sin x+\int 2 x \sin x d x-2 \int \sin x d x+\int 2 x \sin x d x+c_1 x+c\)
y = \(-x^2 \sin x-4 x \cos x+6 \sin x+c_1 x+c\)
2. Solve \(\frac{d^3 y}{d x^3}=x+\log x\)
Solution:
The given differential equation is \(\frac{d^3 y}{d x^3}=x+\log x\)
Integrating w.r.t x once, we get \(\frac{d^2 y}{d x^2}=\frac{x^2}{2}+x \log x-x+c_1\)
Here \(c_1\) is an integrating constant.
Integrating w.r.t x again, we get \(\frac{d y}{d x}=\frac{x^3}{6}-\frac{x^2}{2}+\int x \log x d x+c_1 x+c_2\)
⇒ \(\frac{d y}{d x}=\frac{x^3}{6}-\frac{x^2}{2}-\frac{x}{2}+\frac{x^2}{2} \log x+c_1 x+c_2\), using Integration by parts
Here \(c_2\) is an integrating factor.
Integrating again w.r.t x,
y = \(\frac{x^4}{24}-\frac{x^3}{6}-\frac{x^2}{4}+c_1 \frac{x^2}{2}+c_2 x+\int \frac{x^2}{2} \log x d x\)
y = \(\frac{x^4}{24}-\frac{x^3}{6}-\frac{x^2}{4}-\frac{x^3}{18}+\frac{x^3}{6} \log x+c_1 \frac{x^2}{2}+c_2 x+c_3, c_3\) is an arbitrary constant (using integration parts)
y = \(\frac{x^4}{24}-\frac{2}{9} x^3-\frac{x^2}{4}+\frac{x^3}{6} \log x+c_1 \frac{x^2}{2}+c_2 x+c_3\)
3. Solve \(\frac{d^2 y}{d x^2}=\frac{36}{y^2}, y=8-\frac{d y}{d x}=0 \text { when } x=0\)
Solution:
Given differential equation is \(\frac{d^2 y}{d x^2}=\frac{36}{y^2}\)
Multiplying with \(2 \frac{d y}{d x}\) we get, \(2 \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}=2 \cdot \frac{36}{y^2} \cdot \frac{d y}{d x}\)
Integrating w. r.t x, we get \(\left(\frac{d y}{d x}\right)^2=-\frac{76}{y}+c\)
Given that \(\frac{d y}{d x}=0\) when y=8
0 = \(\frac{-76}{8}+c \Rightarrow c=\frac{76}{8}\)
⇒ \(\left(\frac{d y}{d x}\right)^2=\frac{-76}{y}+\frac{76}{8}=76.8\left(\frac{1}{8}-\frac{1}{4}\right)\)
⇒ \(\left(\frac{d y}{d x}\right)=\sqrt{76} \cdot \sqrt{\frac{1}{8}-\frac{1}{y}}\)
Integrating and applying y=8 when x=0
We get \(\sqrt{y^2-8 y}+8 \sinh ^{-1}\left(\sqrt{\frac{y-8}{8}}\right)=3 x\)
4. Solve \(\frac{d^2 y}{d x^2}=3 \sqrt{y}, y=1, \frac{d y}{d x}=2 \text { when } x=0\)
Solution:
The given differential equation is \(\frac{d^2 y}{d x^2}=3 \sqrt{y}\)…..(1)
Multiplying with \(2 \frac{d y}{d x}\) on both side we get, \(2 \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}=3 \sqrt{y} \cdot 2 \cdot \frac{d y}{d x}\)
Integrating on both sides we get \(\left(\frac{d y}{d x}\right)^2=6 \int \sqrt{y} d y+c_1\)
⇒ \(\left(\frac{d y}{d x}\right)^2=6 \cdot \frac{2}{3} \cdot y^{3 / 2}+c_1 \Rightarrow\left(\frac{d y}{d x}\right)^2=4 y^{3 / 2}+c_1\)…….(2)
Here \(c_1\) is an arbitrary constant.
Given that \(\frac{d y}{d x}=2\) when y=1.
(2)\(^2=4.1+c_1 \Rightarrow c_1=0\)
From (2), \(\left(\frac{d y}{d x}\right)^2=4 y^{3 / 2}\)
⇒ \(\frac{d y}{d x}=2 y^{3 / 4} \Rightarrow \frac{d y}{y^{3 / 4}}=2 d x\)
Integrating \(4 y^{1 / 4}=2 x+c\)……..(3)
Here c is an arbitrary constant.
Given that y=1 when x=0
4=c
From (3), \(4 y^{1 / 4}=2 x+4\) is the solution of the given equation.
Higher Order Linear Differential Equations 3 (Non-Constant Coefficients) Exercise 6(d)
1. Solve \(\frac{d^2 y}{d x^2}=x e^x\)
Solution: \(y=(x-2) e^x+c_1 x+c_2\left(c_1, c_2 \text { are arbitrary constants }\right)\)
2. Solve \(\frac{d^2 y}{d x^2}=2\left(y^3+y\right)\) under the condition y=0,\(\frac{d y}{d x}=1\) when x =0.
Solution: \(y=\tan x\)