Differential Equations Of First Order But Not Of First Degree
Differential Equations First Order But Not First Degree Solvable For P
So far we have discussed differential equations of the first order and of the first degree. In this chapter, we will discuss the differential equations of the first order in which the degree of \(\frac{d y}{d x}\) is not of the first degree. For convenience, we denote \(\frac{d y}{d x}\) by p.
An equation of form f(x,y,p)=0, where p is not of the first degree, is called a differential equation of first order and not of the first degree
An equation of the form \(p^n+P_1(x, y) p^{n-1}+\ldots \ldots \ldots+P_{n-1}(x, y) p+P_n(x, y)=0\) is called the general first order equation of degree n(>1).
These Equations Can Be Divided Into Four Types
- Solvable for p
- Solvable for x
- Solvable for y
- Clairaut’s Equation.
Differential Equations of First Order But Not of First Degree Equations Solvable For p
Let f(x, y, p) = 0 ……………..(1) be the given equation of first order and degree_n(>1).
∴ (1) can be written as \(p^n+\mathrm{P}_1(x, y) p^{n-1}+\ldots \ldots . .+\mathrm{P}_n(x, y)=0\) ……………………(2)
If (2) is solved for p, let n solutions be \(p=f_1(x, y), p=f_2(x, y), \ldots \ldots, p=f_n(x, y)\), ……………………(3)
∴ (2) can be expressed in the form \(\left[p-f_1(x, y)\right]\left[p-f_2(x, y)\right] \ldots \ldots .\left[p-f_n(x, y)\right]=0\) ……………………..(4)
Solving each equation in (3), we get n solutions \(\) for n equations respectively.
Since \(\mathrm{F}_i\left(x, y, c_i\right)=0\) is solution to \(p=f_i(x, y)\) for each i, \(\mathrm{F}_i\left(x, y, c_i\right)=0\) is also solution of (4).
∴ The solution of (1) is \(\mathrm{F}_1\left(x, y, c_1\right) \mathrm{F}_2\left(x, y, c_2\right) \ldots \ldots \mathrm{F}_n\left(x, y, c_n\right)=0\)
Since equation (1) is of the first order, the general solution should have only one arbitrary constant.
Taking \(c_1=c_2=\ldots \ldots=c_n=c\), the general solution of (1) is \(\mathrm{F}_1(x, y, c) \mathrm{F}_2(x, y, c) \ldots \ldots \mathrm{F}_n(x, y, c)=0\)
Examples Of Differential Equations Solvable For P
Differential Equations of First Order But Not of First Degree Solved Problems
1. Solve \(p^2-5 p+6=0\)
Solution.
Given equation is \(p^2-5 p+6=0\) ………………..(1)
Solving for p => (p – 3) (p – 2) = 0
=> p = 3 ………………………..(2) and
P = 2 …………………….(3).
Solving (2) and (3), we get: \(\frac{d y}{d x}=3 \Rightarrow \int d y=3 \int d x+c \Rightarrow y=3 x+c \Rightarrow y-3 x-c=0\)
∴ \(\frac{d y}{d x}=2 \Rightarrow \int d y=2 \int d x+c \Rightarrow y=2 x+c \Rightarrow y-2 x-c=0\)
∴ The general solution of (1) is (y – 3x – c)(y – 2x – c) = 0
Solved Problems On Equations Solvable For P In Differential Equations
2. Solve \(x+y p^2=(1+x y) p\)
Solution.
Given \(x+y p^2=(1+x y) p\) ………………….(1)
Solving for \(\mathrm{p} \Rightarrow y p^2-p-x y p+x=0\)
⇒ \(p(y p-1)-x(y p-1)=0 \Rightarrow(p-x)(y p-1)=0 \Rightarrow p=x\) …………………(2)
yp -1 = 0 ……………………(3)
Solving (2) and (3): \(p=x \Rightarrow \frac{d y}{d x}=x \Rightarrow \int d y=\int x d x+c \Rightarrow y=\frac{x^2}{2}+c\)
⇒ \(2 y-x^2-2 c=0 \text { and } p=\frac{1}{y} \Rightarrow \frac{d y}{d x}=\frac{1}{y} \Rightarrow \int y d y=\int d x+c \Rightarrow\left(y^2 / 2\right)\)
= \(x+c \Rightarrow y^2-2 x-2 c=0\)
∴ The general solution of (1) is \(\left(2 y-x^2-2 c\right)\left(y^2-2 x-2 c\right)=0\)
Step-By-Step Solutions For First Order But Not First Degree Equations For P
3. Solve \(4 y^2 p^2+2 x y(3 x+1) p+3 x^3=0\)
Solution.
Given equation is \(4 y^2 p^2+2 x y(3 x+1) p+3 x^3=0\) ……………………….(1)
solving for p, we have : \(4 y^2 p^2+6 x^2 y p+2 x y p+3 x^3=0\)
⇒ \(2 y p\left(2 y p+3 x^2\right)+x\left(2 y p+3 x^2\right)=0 \quad \Rightarrow\left(2 y p+3 x^2\right)(2 y p+x)=0\)
⇒ \(p=-3 x^2 / 2 y\) ………………….(2)
p = -x/2y …………………….(3)
Solving(2): \(2 y d y=-3 x^2 d x \Rightarrow\) Integrating, \(y^2+x^3+c=0\)
Solving (3) : \(2 y d y=-x d x \Rightarrow\) Integrating, \(2 y^2+x^2+c=0\).
Thus the solutions of (2) and (3) are \(y^2+x^3+c=0 \text { and } 2 y^2+x^2+c=0\)
∴ The general solution of (1) is \(\left(y^2+x^3+c\right)\left(2 y^2+x^2+c\right)=0\)
Example. 1 Solve \(x y p^2+\left(x^2+x y+y^2\right) p+\left(x^2+x y\right)=0\)
Solution.
Given equation is \(x y p^2+\left(x^2+x y+y^2\right) p+\left(x^2+x y\right)=0\) ……………………(1)
⇒ \(x y p^2+x^2 \cdot p+x y p+x^2+y^2 p+x y=0 \Rightarrow x p(y p+x)+x(y p+x)+y(y p+x)=0\)
⇒ \((y p+x)(x p+x+y)=0 \quad \Rightarrow y p+x=0\) ……………………(2)
x p+x+y=0 ……………………..(3)
Solving (2): \(x \frac{d y}{d x}+x=0 \Rightarrow y d y+x d x=0\)
Integrating, \(y^2 / 2+x^2 / 2=c_1 \Rightarrow x^2+y^2=c\)
Solving (3): \(x \frac{d y}{d x}+x+y=0 \Rightarrow x d y+y d x+x d x=0 \Rightarrow d(x y)+x d x=0\)
Integrating: \(x y+\left(x^2 / 2\right)=c \Rightarrow 2 x y+x^2=2 c\)
Thus the solutions of (2) and (3) are \(x^2+y^2-c=0\) and \(2 x y+x^2-2 c=0\)
∴ The general solution of (1) is \(\left(x^2+y^2-c\right)\left(2 x y+x^2-2 c\right)=0\)
Applications Of Equations Solvable For p In Mathematics
Example. 2 Solve \(p^2+2 p y \cot x=y^2\)
Solution.
Given equation is \(p^2+(2 y \cot x) p=y^2\) ………………………(1)
Solving for \(p:(p+y \cot x)^2=y^2\left(1+\cot ^2 x\right)=y^2 \ cosec^2 x \Rightarrow p+y \cot x=\pm y \ cosec x\)
⇒ \((p+y \cot x+y \ cosec x)(p+y \cot x-y \ cosec x)=0\)
⇒ \( ……………………(2) and
⇒ [latex]\frac{d y}{d x}+y(\cot x-\ cosec x)=0\) …………………….(3)
In (2) and (3), variables are separable. Solving (2):
⇒ \(\frac{d y}{y}=-(\ cosec x+\cot x) d x=-\frac{1+\cos x}{\sin x} d x=-\cot (x / 2) d x\)
Integrating, \(\log y=-2 \log \sin (x / 2)+\log c \Rightarrow y=c \ cosec^2(x / 2)\)
Solving (3) : \(\frac{d y}{y}=(\ cosec x-\cot x) d x=\frac{1-\cos x}{\sin x} d x=\tan (x / 2) d x\)
Integrating, \(\log y=2 \log \sec (x / 2)+\log c \Rightarrow y=c \sec ^2(x / 2)\)
Thus the solutions of (2) and (3) are \(y-c \ cosec^2(x / 2)=0 \text { and } y-\sec ^2(x / 2)=0\)
∴ The general solution of (1) is \(\left[y-c \ cosec^2(x / 2)\right]\left[y-c \sec ^2(x / 2)\right]=0\)
Solving Non-First Degree Differential Equations For P With Examples
Example. 3 \(x^2\left(\frac{d y}{d x}\right)^2-2 x y \frac{d y}{d x}+2 y^2-x^2=0\)
Solution.
Given \(x^2 p^2-2 x y p+\left(2 y^2-x^2\right)=0 \text { where } \frac{d y}{d x}=p\) …………………….(1)
Solving (1) for p: \(\Rightarrow p=\frac{2 x y \pm \sqrt{4 x^2 y^2-4 x^2\left(2 y^2-x^2\right)}}{2 x^2}=\frac{y \pm \sqrt{x^2-y^2}}{x}\)
⇒ \(p=\frac{y+\sqrt{x^2-y^2}}{x}\) …………………….(2)
p \(=\frac{y-\sqrt{x^2-y^2}}{x}\) …………………………(3)
(2) and (3) are homogeneous equations.
Put \(y=v x \Rightarrow \frac{d y}{d x}=p=v+x \frac{d v}{d x}\) …………………….(4)
(2) and (4) \(\Rightarrow v+x \frac{d v}{d x}=\frac{v x+x \sqrt{1-v^2}}{x}=v+\sqrt{1-v^2} \Rightarrow x \frac{d v}{d x}=\sqrt{1-v^2} \Rightarrow \frac{d v}{\sqrt{1-v^2}}=\frac{d x}{x}\)
⇒ \(\int \frac{d v}{\sqrt{1-v^2}}=\int \frac{d x}{x}+\log c \Rightarrow \sin ^{-1} v=\log x+\log c=\log c x\)
The solution of (2) is \(\sin ^{-1}(y / x)-\log c x=0\)
Similarly, the solution of (3) is \(\sin ^{-1}(y / x)+\log c x=0\)
∴ The general solution of (1) is \(\left[\sin ^{-1}(y / x)-\log c x\right]\left[\sin ^{-1}(y / x)+\log c x\right]=0\)
Methods To Solve Equations Of First Order But Not First Degree Solvable For P
Example. 4 Solve \(p^3+\left(2 x-y^2\right) p^2=2 x y^2 p\)
Solution.
Given equation is \(p^3+\left(2 x-y^2\right) p^2-2 x y^2 p=0\) …………………….(1)
⇒ \(p\left[p^2+\left(2 x-y^2\right) p-2 x y^2\right]=0 \Rightarrow p\left(p^2+2 x p-y^2 p-2 x y^2\right)=0\)
⇒ \(p(p+2 x)\left(p-y^2\right)=0 \Rightarrow p=0\) ……………………(2)
p+2 x=0 …………………..(3)
∴ \(p-y^2=0\) …………………..(4)
Solving (2): \(p=0 \Rightarrow \frac{d y}{d x}=0 \Rightarrow y=c\)
Solving (3): \(\frac{d y}{d x}=-2 x \Rightarrow d y=-2 x d x \Rightarrow \int d y=-2 \int x d x+c\)
⇒ \(y=-2\left(x^2 / 2\right)+c \Rightarrow y+x^2=c\)
Solving (4) : \(\frac{d y}{y^2}=d x \Rightarrow \int \frac{d y}{y^2}=\int d x+c \Rightarrow-\frac{1}{y}=x+c \Rightarrow x y+c y+1=0\)
∴ The G.S. of (1) is \((y-c)\left(y+x^2-c\right)(x y+c y+1)=0\)
Differential Equations Solvable For P Examples And Properties
Example. 5 Solve \(p^4-(x+2 y+1) p^3+(x+2 y+2 x y) p^2-2 x y p=0\)
Solution.
Given \(p\left[p^3-(x+2 y+1) p^2+(x+2 y+2 x y) p-2 x y\right]=0\) …………………….(1)
⇒ \(p\left[p^3-(x+2 y) p^2-p^2+(x+2 y) p+2 x y p-2 x y\right]=0\)
⇒ \(p\left[\left(p^3-p^2\right)-(x+2 y) p(p-1)+2 x y(p-1)\right]=0\)
⇒ \(p(p-1)\left[p^2-(x+2 y) p+2 x y\right]=0\)
⇒ p = 0 …………………..(2)
⇒ p – 1 = 0 …………………….(3)
⇒ \(p^2-(x+2 y) p+2 x y=0\) ……………………….(4)
Solving (2) : \(\frac{d y}{d x}=0 \Rightarrow d y=0. d x \Rightarrow \int d y=0+c \Rightarrow y=c\)
Solving (3): \(p-1=0 \Rightarrow \frac{d y}{d x}=1 \Rightarrow \int d y=\int d x+c \Rightarrow y=x+c\)
Solving (4): \((p-x)(p-2 y)=0 \Rightarrow p=x, p=2 y\)
p = \(x \Rightarrow \frac{d y}{d x}=x \Rightarrow \int d y=\int x d x+c \Rightarrow y=\frac{x^2}{2}+c\)
p = \(2 y \Rightarrow \frac{d y}{d x}=2 y \Rightarrow \int \frac{d y}{y}=\int 2 d x+c \Rightarrow \log y=2 x+c \Rightarrow y=c^{2 x+c}\)
∴ The general solution of (1) is \((y-c)(y-x-c)\left(y-\frac{x^2}{2}-c\right)\left(y-e^{2 x+c}\right)=0\)
Worked on Examples Of Differential Equations Solvable For P
Example. 6 Solve \(x y p^2+p\left(3 x^2-2 y^2\right)-6 x y=0\)
Solution.
Given \(x y p^2+p\left(3 x^2-2 y^2\right)-6 x y=0\) ……………………(1)
x \(p(y p+3 x)-2 y(y p+3 x)=0 \Rightarrow(y p+3 x)(x p-2 y)=0 \Rightarrow p=2 y / x, p=-3 x / y\)
Solving: \(p=\frac{2 y}{x} \Rightarrow \frac{d y}{d x}=\frac{2 y}{x} \Rightarrow \int \frac{d y}{y}=2 \int \frac{d x}{x}+\log c\)
⇒ \(\log y=2 \log x+\log c=\log c x^2 \Rightarrow y=c x^2\)
Solving: \(p=-\frac{3 x}{y} \Rightarrow \frac{d y}{d x}=-\frac{3 x}{y} \Rightarrow \int y d y=-\int 3 x d x+c \Rightarrow \frac{y^2}{2}=-\frac{3 x^2}{2}+\frac{c}{2} \Rightarrow y^2+3 x^2=c\)
Hence the general solution of (1) is \(\left(y-c x^2\right)\left(y^2+3 x^2-c\right)=0\)