Differential Equations of First Order and First Degree Solved Example Problems Exercise 2.3
Example. 1. Solve \(y\left(x y+2 x^2 y^2\right) d x+x\left(x y-x^2 y^2\right) dy=0\)
Solution.
Given \(y\left(x y+2 x^2 y^2\right) d x+x\left(x y-x^2 y^2\right) d y=0\) ………………….(1)
Comparing (1) with Mdx + Ndy = 0 => \(\mathrm{M}=y\left(x y+2 x^2 y^2\right), \mathrm{N}=x\left(x y-x^2 y^2\right)\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=2 x y+6 x^2 y^2, \frac{\partial \mathrm{N}}{\partial x}=2 x y-3 x^2 y^2 \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\)
(1) is not an exact equation. But (1) is of the form y f(xy) dx + x g(xy) dy = 0
Also \(\mathrm{M} x-\mathrm{N} y=x^2 y^2+2 x^3 y^3-x^2 y^3+x^3 y^3=3 x^3 y^3 \neq 0\)
∴ I.F. = \(\frac{1}{M x-N y}=\frac{1}{3 x^3 y^3}\)
Multiplying (1) by \(\frac{1}{3 x^3 y^3}\)
⇒ \(\frac{x y^2+2 x^2 y^3}{3 x^3 y^3} d x+\frac{x^2 y-x^3 y^2}{3 x^3 y^3} d y=0\) …………………(2) which of the form
where \(\mathrm{M}_1=\frac{x y^2+2 x^2 y^3}{3 x^3 y^3}=\frac{1}{3 x^2 y}+\frac{2}{3 x} \text { and } \mathrm{N}_1=\frac{x^2 y-x^3 y^2}{3 x^3 y^3}=\frac{1}{3 x y^2}-\frac{1}{3 y}\)
⇒ \(\frac{\partial \mathrm{M}_1}{\partial y}=-\frac{1}{3 x^2 y^2}, \frac{\partial \mathrm{N}_1}{\partial x}=-\frac{1}{3 x^2 y^2} \Rightarrow \frac{\partial \mathrm{M}_1}{\partial y}=\frac{\partial \mathrm{N}_1}{\partial x} \Rightarrow\) (2) is exact
Now (2) can be solved by rearranging terms: \(\frac{1}{x^2 y} d x+\frac{2}{x} d x+\frac{1}{x y^2} d y-\frac{1}{y} d y=0\)
⇒ \(\frac{y d x+x d y}{x^2 y^2}+2 \cdot \frac{1}{x} d x-\frac{1}{y} d y=0 \Rightarrow \frac{d(x y)}{x^2 y^2}+\frac{2}{x} d x-\frac{1}{y} d y=0\)
⇒ \(\int \frac{d(x y)}{(x y)^2}+2 \int \frac{1}{x} d x-\int \frac{1}{y} d y=c \Rightarrow-\frac{1}{x y}+2 \log |x|-\log |y|=c \text { is the G.S. of (1) }\)
Differential Equations Of First Order And First Degree Exercise 2.3
Example. 2. Solve \((x y \sin x y+\cos x y) y d x+(x y \sin x y-\cos x y) x d y=0\)
Solution.
Given \((x y \sin x y+\cos x y) y d x+(x y \sin x y-\cos x y) x d y=0\) …………………. (1)
where \(\mathrm{M}=x y^2 \sin x y+y \cos x y, \mathrm{~N}=x^2 y \sin x y-x \cos x y\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=\left(x^2 y^2+1\right) \cos x y+x y \sin x y, \frac{\partial \mathrm{N}}{\partial x}=3 x y \sin x y+\left(x^2 y^2-1\right) \cos x y\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\)
∴ (1) is not an exact equation
But (1) is of the form \(y f(x y) d x+x g(x y) d y=0\)
Also \(\mathrm{M} x-\mathrm{Ny}=x^2 y^2 \sin x y+x y \cos x y-x^2 y^2 \sin x y+x y \cos x y=2 x y \cos x y \neq 0\)
∴ I.F = \(\frac{1}{\mathrm{M} x-\mathrm{N} y}=\frac{1}{2 x y \cos x y}\)
Multiplying (1) with \(\frac{1}{(2 x y \cos x y)}\)
⇒ \(\frac{1}{2}\left(y \tan x y+\frac{1}{x}\right) d x+\frac{1}{2}\left(x \tan x y-\frac{1}{y}\right) d y=0\)……………………(2)
(2) is of the form \(\mathrm{M}_1 d x+\mathrm{N}_1 d y=0\) where \(\mathrm{M}_1=\frac{1}{2}\left(y \tan x y+\frac{1}{x}\right)\)
and \(\mathrm{N}_1=\frac{1}{2}\left(x \tan x y-\frac{1}{y}\right)\)
⇒ \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{1}{2}\left(x y \sec ^2 x y+\tan x y\right)=\frac{\partial \mathrm{N}_{\mathrm{l}}}{\partial x}\)
∴ (2) is an exact equation
∴ The general solution of (2) is \(\int^x \mathrm{M}_1 d x+\int\) ( terms of \(\mathrm{N}_1\) not containing x) dy = C
⇒ \(\int^x\left(\frac{1}{2} y \tan x y+\frac{1}{2 x}\right) d x+\int\left(-\frac{1}{2 y}\right) d y=c\)
⇒ \(\frac{1}{2} y \cdot \frac{\log |\sec x y|}{y}+\frac{1}{2} \log |x|-\frac{1}{2} \log |y|=\frac{1}{2} \log c\)
⇒ \(\log \left|\frac{x}{y} \sec x y\right|=\log c \Rightarrow x \sec x y=c y\)
Homogeneous Equations Solved Problems Exercise 2.3
Example 3. Solve \(\left(x^3 y^3+x^2 y^2+x y+1\right) y d x+\left(x^3 y^3-x^2 y^2-x y+1\right) x d y=0\)
Solution.
Given equation is of the form \(y f(x y) d x+x g(x y) d y=0\) and is not an exact equation (verify),
where M = \(x^3 y^4+x^2 y^3+x y^2+y, \mathrm{~N}=x^4 y^3-x^3 y^2-x^2 y+x\)
Also \(\mathrm{M} x-\mathrm{N} y=x^4 y^4+x^3 y^3+x^2 y^2+x y-x^4 y^4+x^3 y^3+x^2 y^2-x y\)
= \(2 x^3 y^3+2 x^2 y^2=2 x^2 y^2(x y+1) \neq 0\)
I.F = \(\frac{1}{\mathrm{M} x-\mathrm{N} y}=\frac{1}{2 x^2 y^2(x y+1)}\)
The given equation can be written as \((x y+1)\left(x^2 y^2+1\right) y d x+(x y-1)\left(x^2 y^2-1\right) x d y=0\) ……………………(1)
Multiplying(1) with \(\frac{1}{2 x^2 y^2(x y+1)} \Rightarrow \frac{x^2 y^2+1}{2 x^2 y} d x+\frac{(x y-1)^2}{2 x y^2} d y=0\) ……………………..(2)
(2) is exact (verify) where \(\mathrm{M}_1=\frac{x^2 y^2+1}{2 x^2 y}=\frac{y}{2}+\frac{1}{2 x^2 y}\)
and \(\mathrm{N}_1=\frac{(x y-1)^2}{2 x y^2}=\frac{x^2 y^2+1-2 x y}{2 x y^2}=\frac{x}{2}+\frac{1}{2 x y^2}-\frac{1}{y}\)
∴ \(\text { The G.S. of (2) is } \int^x\left(\frac{y}{2}+\frac{1}{2 x^2 y}\right) d x+\int\left(-\frac{1}{y}\right) d y=c \Rightarrow \frac{x y}{2}+\frac{1}{2 y}\left(-\frac{1}{x}\right)-\log |y|=c\)
⇒ \(x y-\frac{1}{x y}-\log y^2=2 c\)