Differential Equations of First Order But Not of First Degree Equations Homogeneous In X and Y
When the given equation is homogeneous in x and y it can be written as \(\mathrm{F}\left(\frac{d y}{d x}, \frac{y}{x}\right)=0\).
It is then possible to solve it for \(p=\frac{d y}{d x}\).
Differentiating with respect to x, we get \(p=f(p)+x f^{\prime}(p) \frac{d p}{d x}\) from which we can separate the variables and write \(\frac{d x}{x}=\frac{f^{\prime}(p) d p}{p-f(p)}\) and solve
Differential Equations of First Order But Not of First Degree Solved Problems
Example. 1. Solve the D.E. \(y^2+x y p-x^2 p^2=0\) which is homogeneous in x and y.
Solution.
Given equation is \(y^2+x y p-x^2 p^2=0\) …………………… (1)
Dividing(1)by \(x^2\), we have \(\frac{y^2}{x^2}+\frac{y}{x} p-p^2=0\)
⇒ \(\frac{y}{x}=\frac{-p \pm \sqrt{p^2+4 p^2}}{2}=\frac{1 \pm \sqrt{5}}{2} p \Rightarrow \frac{y}{x}=c \cdot \frac{d y}{d x}\) where c = \(\frac{-1 \pm \sqrt{5}}{2}\)
⇒ \(\frac{d x}{x}=c \cdot \frac{d y}{y}\)
∴ Integrating \(\log x=c \log y+\log c_1\) (where \(c_1\) is a constant)
⇒ \( \log x=\log \left(c_1 y^c\right) \Rightarrow x=c_1 y^c \Rightarrow x-c_1 y^c=0\)
∴ The required general solution of equation (1) is
⇒ \(\left[x-c_1 y^{\left(\frac{-1+\sqrt{5}}{2}\right)}\right]\left[x-c_1 y^{\left(\frac{-1-\sqrt{5}}{2}\right)}\right]=0\)
∴ \(c=\frac{-1 \pm \sqrt{5}}{2}\)
Differential Equations Of First Order But Not Of First Degree Examples
Example. 2. Solve \(8 y^2-4 x y p-x^2 p^2=0\)
Solution.
Given equation is \(8 y^2-4 x y p-x^2 p^2=0\)
Dividing with \(x^2, 8(y / x)^2-4(y / x) p-p^2=0\)
Solving for (y / x), we get \(\frac{y}{x}=\frac{4 p \pm \sqrt{16 p^2+32 p^2}}{2}=(2 \pm 2 \sqrt{3}) p \Rightarrow \frac{y}{x}=(2 \pm 2 \sqrt{3}) \frac{d y}{d x}\)
Separating the variables, \(\frac{d y}{y}=\frac{1}{2 \pm 2 \sqrt{3}} \cdot \frac{d x}{x}\)
Integrating on both sides,
log y \(=\frac{1}{2(1+\sqrt{3})} \log x+\log c \text { and } \log y=\frac{1}{2(1-\sqrt{3})} \log x+\log c\)
⇒ \(y=c x^{1 / 2(1+\sqrt{3})} \text { and } y=c x^{1 / 2(1-\sqrt{3})}\)
C.S. is \(\left[y-c x^{1 / 2(1+\sqrt{3})}\right]\left[y-c x^{1 / 2(1-\sqrt{3})}\right]=0\)
Solved Problems On Homogeneous Equations In x And y
Example. 3. Solve \(y=y p^2+2 p x\)
Solution.
Given equation is \(y=y p^2+2 p x \text { i.e. } y\left(1-p^2\right)=2 p x . \quad y=\frac{2 p x}{1-p^2}\) ………………………(1)
Differentiating w.r.t. x, \(\frac{2 d p}{p\left(1-p^2\right)}+\frac{d x}{x}=0\)
=> Using partial fractions, \(\left(\frac{2}{p}+\frac{1}{1-p}-\frac{1}{1+p}\right) d p+\frac{d x}{x}=0\)
Integrating, \(x=\frac{c\left(1-p^2\right)}{p^2}\) …………………………..(2)
Eliminating p between (1) and (2) we get, \(y^2=4 x c+4 c^2\) which is the required solution.
Differential Equations of First Order But Not of First Degree Clairaut’s Equation
The differential equation of the form y = xp + f (p) is called Clairaut’s equation.
This equation is solved by considering it as y = f(x,p), solvable for the y type.
Solution of Clairaut’s Equation :
Let the given equation be y = xp + f (p) ……………………..(1)
Differentiating (1) w.r.t. x: \(\Rightarrow \frac{d y}{d x}=x \frac{d p}{d x}+p+f^{\prime}(p) \frac{d p}{d x}\)
⇒ \(p=\left[x+f^{\prime}(p)\right] \frac{d p}{d x}+p \Rightarrow\left[x+f^{\prime}(p)\right] \frac{d p}{d x}=0\)
⇒ \(\frac{d p}{d x}=0\) ……………………(2)
x\(+f^{\prime}(p)=0\) …………………….(3)
Solving (2): \(\frac{d p}{d x}=0 \Rightarrow p=c\) where c is a real number. ………………………..(4)
Eliminating p from (1) and (4) y = cx + f(c).
∴ The general solution of Clairaut’s equation (1) is y = cx + f(c).
Note. Solving the equation (3): \(x+f^{\prime}(p)=0\) ……………………….(5)
By eliminating p from (1) and (5) we get the solution Φ (x,y) = 0.
As this solution does not contain any arbitrary constant this is not the general solution. This solution is called the singular solution of (1).
Differential Equations of First Order But Not of First Degree Working Rule for Solving Clairaut’s Equation :
1. The given equation can be written in the form y = xp + f(p) ………………..(a)
2. In order to find the solution of (a), replace p by c where c is any real number.
3. Putting p = c in (a), we get y = xc + f(c).
∴ The general solution of (a) is y = xc + f{c).
Differential Equations of First Order But Not of First Degree Solved Problems
Example. 1. Solve \((y-x p)(p-1)=p\)
Solution.
Given equation is \((y-x p)(p-1)=p\) …………………….(1)
⇒ \(y(p-1)-x p(p-1)=p \Rightarrow y(p-1)=x p(p-1)+p\)
⇒ \(y=x p+\frac{p}{p-1}\) which is in Clairaut’s form.
The genera] solution of (1) is \(y=x c+\frac{c}{c-1}\) where c is any real number
Example. 2. Solve \(p=\tan (x p-y)\)
Solution:
Given equation is \(p=\tan (x p-y)\)
⇒ x p-y=\text{Tan}^{-1} p \Rightarrow y=x p-\text{Tan}^{-1} p[/latex] which is in Clairaut’s form.
∴ The general solution of (1) is \(y=x c-\text{Tan}^{-1} c\) where c is any real number.
Problems On First-Order But Not First-Degree Equations Homogeneous In x And y
Example. 3. Solve \(y^2-2 p x y+p^2\left(x^2-1\right)=m^2\)
Solution.
Given equation is \(y^2-2 p x y+p^2\left(x^2-1\right)=m^2\) ……………………….(1)
⇒ \(y^2-2 p x y+p^2 x^2=p^2+m^2 \Rightarrow(y-p x)^2=p^2+m^2\)
⇒ \(y-p x=\sqrt{p^2+m^2} \Rightarrow y=x p+\sqrt{p^2+m^2}\) which is in Clairaut s form.
∴ The general solution of (1) is \(y=x c+\sqrt{c^2+m^2}\)
Applications Of Clairaut’s Equation In Differential Equations.
Example. 4. Solve \(\sin p x \cos y=\cos p x \sin y+p\)
Solution.
Given equation is \(\sin p x \cos y-\cos p x \sin y=p\) …………………..(1)
⇒ \(\sin (p x-y)=p \Rightarrow p x-y=\sin ^{-1} p \Rightarrow y=p x-\sin ^{-1} p\) which is in Clairaut’s form.
∴ The general solution of (1) is \(y=c x-\sin ^{-1} c\), c being an arbitrary real number.
Solved Problems On Homogeneous Equations Of First Order Not First Degree
Example. 5. Solve \(p=\log (p x-y)\)
Solution.
Given equation is \(p=\log (p x-y)\) ……………….. (1)
⇒ \(p x-y=e^p \Rightarrow y=p x-e^p\) which is in Clairaut’s form.
∴ The general solution of (1) is \(y=c x-e^c\) where c is any real number.