Differential Equations of First Order and First Degree Solved Example Problems Exercise 2.2
Example. 1. Solve \(x^2 y d x-\left(x^3+y^3\right) d y=0\)
Solution.
Given equation is \(x^2 y d x-\left(x^3+y^3\right) d y=0\) ………………………(1)
Comparing (1) with \(\mathrm{M} d x+\mathrm{N} d y=0 \Rightarrow \mathrm{M}=x^2 y, \mathrm{~N}=-\left(x^3+y^3\right)\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=x^2, \frac{\partial \mathrm{N}}{\partial x}=-3 x^2 \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\) => (1) is not exact equation
But (1) is a homogeneous equation in x and y.
Now \(\mathrm{M} x+\mathrm{N} y=x^3 y-x^3 y-y^4=-y^4 \neq 0\)
∴ I.F. = Integrating Factor = \(=\frac{1}{\mathrm{M} x+\mathrm{N} y}=-\frac{1}{y^4}\)
Multiplying(1)by \(\left(-1 / y^4\right) \Rightarrow \frac{-x^2}{y^3} d x+\left(\frac{x^3}{y^4}+\frac{1}{y}\right) d y=0\) ………………………..(2)
(2) is in the form \(\mathrm{M}_1 d x+\mathrm{N}_1 d y=0\) where \(\mathrm{M}_1=-\frac{x^2}{y^3} \text { and } \mathrm{N}_1=\frac{x^3}{y^4}+\frac{1}{y}\)
since \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{3 \dot{x}^2}{v^4}=\frac{\partial \mathrm{N}_1}{\partial x}\), (2) is an exact equation.
(1) Integrating MjW.r.t. x treating y as constant
⇒ \(\int^x \mathrm{M}_1 d x=\int\left(-x^2 / y^3\right) d x=-x^3 / 3 y^3\) …………………….(3)
(2) \(\int\)(terms of \(\mathrm{N}_1\) not containing x) dy = \(d y=\int \frac{1}{y} d y=\log |y|\) ……………………….(4)
The GS. of (2) is (3) + (4) = c=> \(\Rightarrow-\frac{x^3}{3 y^3}+\log |y|=c\)
∴ The G.S of (1) is \(\log \left|\frac{y}{c}\right|=\frac{x^3}{3 y^3} \Rightarrow y=c e^{\left(x^3 / 3 y^3\right)}\)
Example 2. Solve \(y^2 d x+\left(x^2-x y-y^2\right) d y=0\)
Solution:
Given equation is \(y^2 d x+\left(x^2-x y-y^2\right) d y=0\)
comparing(1) with \(\mathrm{M} d x+\mathrm{N} d y=0 \Rightarrow \mathrm{M}=y^2, \mathrm{~N}=x^2-x y-y^2 \Rightarrow \frac{\partial \mathrm{M}}{\partial y}=2 y, \frac{\partial \mathrm{N}}{\partial x}=2 x-y\)
Since \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\), (1) is not exact. But (1) is a homogeneous equation in x andy.
⇒ \(\mathrm{M} x+\mathrm{N} y=x y^2+x^2 y-x y^2-y^3=y\left(x^2-y^2\right) \neq 0\)
∴ \(\mathrm{I} . \mathrm{F}=\frac{1}{\mathrm{M} x+\mathrm{N} y}=\frac{1}{y\left(x^2-y^2\right)}\)
Multiplying (1) by \(\frac{1}{y\left(x^2-y^2\right)} \Rightarrow \frac{y}{x^2-y^2} d x+\frac{x^2-x y-y^2}{y\left(x^2-y^2\right)} d y=0\) …………………….(2)
(2) is in the form \(\mathrm{M}_1 d x+\mathrm{N}_1 d y=0\) where \(\mathrm{M}_1=\frac{y}{x^2-y^2}\)
and \(\mathrm{N}_1=\frac{x^2-x y-y^2}{y\left(x^2-y^2\right)}=\frac{x^2-y^2}{y\left(x^2-y^2\right)}-\frac{x y}{y\left(x^2-y^2\right)}=\frac{1}{y}-\frac{x}{x^2-y^2}\)
⇒ \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{x^2+y^2}{\left(x^2-y^2\right)^2}=\frac{\partial \mathrm{N}_1}{\partial x}\), (2) is an exact equation
(1) Integrating \(\mathrm{M}_1\) treating y as constant
⇒ \(\int^x \mathrm{M}_1 d x=\int^x \frac{y}{x^2-y^2} d x=y \cdot \frac{1}{2 y} \log \left|\frac{x-y}{x+y}\right|=\frac{1}{2} \log \left|\frac{x-y}{x+y}\right|\) …………………..(3)
(2) \(\int\) (terms of \(\mathrm{N}_1\) not containing x) dy = \(\int \frac{1}{y} d y=\log |y|\) …………………….(4)
∴ The general solution of (2) => G.S. of (1) is (3) + (4) = c.
⇒ \(\frac{1}{2} \log \left|\frac{x-y}{x+y}\right|+\log |y|=\log c\)
⇒ \(\log y \sqrt{\frac{x-y}{x+y}}=\log c \Rightarrow y \sqrt{\frac{x-y}{x+y}}=c \Rightarrow y^2(x-y)=c^2(x+y)\)
Example. 3. Solve \(y^2+x^2 \frac{d y}{d x}=x y \frac{d y}{d x}\)
Solution.
Given \(y^2 d x+\left(x^2-x y\right) d y=0\) ……………………(1)
Comparing (1) with \(\mathrm{M} d x+\mathrm{N} d y=0 \Rightarrow \mathrm{M}=y^2, \mathrm{~N}=x^2-x y\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=2 y, \frac{\partial \mathrm{N}}{\partial x}=2 x-y \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\) => (1) is not an exact equation.
But (1) is a homogeneous equation in x and y.
I.F = \(\frac{1}{\mathrm{M} x+\mathrm{N} y}=\frac{1}{x y^2+x^2 y-x y^2}=\frac{1}{x^2 y}(\mathrm{M} x+\mathrm{N} y \neq 0)\)
Multiplying (1) by \(\frac{1}{x^2 y} \text {, we get } \frac{y^2}{x^2 y} d x+\frac{x^2-x y}{\dot{x}^2 y} d y=0 \Rightarrow \frac{y}{x^2} d x+\left(\frac{1}{y}-\frac{1}{x}\right) d y=0\) …………………….(2)
(2) is in the form \(\mathrm{M}_1 d x+\mathrm{N}_1 d y=0 \text { where } \mathrm{M}_1=\frac{y}{x^2} \text { and } \mathrm{N}_1=\frac{1}{y}-\frac{1}{x}\)
since \(\frac{\partial \mathrm{M}_1}{\partial y}=\frac{1}{x^2}=\frac{\partial \mathrm{N}_1}{\partial x}\); (2) is an exact equation.
(1) Integrating \(M_1\) w.r.t x, treating y as constant: \(\int \mathrm{M}_1 d x=\int\left(y / x^2\right) d x=-y / x\) …………………….(3)
(2) \(\int\) (terms of \(\mathrm{N}_1\) not containing x) dy = \(\int(1 / y) d y=\log y\) …………………….(4)
The G.S. of (2) is (3) + (4) = c => -(y/x) + logy = -log c
∴ The G.S. of (1) is log c + logy = y/x => y/x = log(cy)