Differential Equations of First Order and First Degree Solved Example Problems Exercise 2.4
Example 1. Solve \(\left(y+\frac{y^3}{3}+\frac{x^2}{2}\right) d x+\frac{1}{4}\left(x+x y^2\right) d y=0\)
Solution:
Given \(\left(y+\frac{y^3}{3}+\frac{x^2}{2}\right) d x+\frac{1}{4}\left(x+x y^2\right) d y=0\) …………………(1)
where \(\mathrm{M}=y+\frac{y^3}{3}+\frac{x^2}{2}, \mathrm{~N}=\frac{1}{4}\left(x+x y^2\right)\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=1+y^2, \frac{\partial \mathrm{N}}{\partial x}\)
= \(\frac{1}{4}\left(1+y^2\right) \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x} \Rightarrow\) (1) is not exact
Now \(\frac{1}{\mathrm{~N}}\left(\frac{\partial \mathrm{M}}{\partial y}-\frac{\partial \mathrm{N}}{\partial x}\right)\)
= \(\frac{4}{x\left(1+y^2\right)}\left[\left(1+y^2\right)-\frac{1}{4}\left(1+y^2\right)\right]\)
= \(\frac{4}{x\left(1+y^2\right)} \cdot \frac{3}{4}\left(1+y^2\right)=\frac{3}{x}=f(x)\)
∴ I.F \(=\exp \left[\int\left(\frac{3}{x}\right) d x\right]=\exp (3 \log x)=e^{\log x^3}=x^3\)
Multiplying (1) by \(x^3 \Rightarrow\left(y+\frac{y^3}{3}+\frac{x^2}{2}\right) x^3 d x+\frac{1}{4}\left(x+x y^2\right) x^3 d y=0\) …………………(2)
Now (2) is an exact equation (verify) where \(\mathrm{M}_1=x^3 y+\frac{x^3 y^3}{3}+\frac{x^5}{2}, \mathrm{~N}_1=\frac{1}{4} x^4+\frac{1}{4} x^4 y^2\)
The general solution (2) is \(\int^x M_1 d x+\int\left(\text { terms of } \mathrm{N}_1 \text { not containing } x\right) d y=c\)
⇒ \(\int^x\left(x^3 y+\frac{1}{3} x^3 y^3+\frac{1}{2} x^5\right) d x+\int 0 d y=c_1\)
⇒ \(\frac{x^4 y}{4}+\frac{x^4 y^3}{12}+\frac{x^6}{12}=\frac{c}{12} \Rightarrow 3 x^4 y+x^4 y^3+x^6=c\)
Differential Equations Of First Order And First Degree Exercise 2.4
Example 2. Solve \(\left(x^2+y^2+2 x\right) d x+2 y d y=0\)
Solution.
Given equation is \(\left(x^2+y^2+2 x\right) d x+2 y d y=0\) ……………………..(1)
Where \(\mathrm{M}=x^2+y^2+2 x, \mathrm{~N}=2 y \Rightarrow \frac{\partial \mathrm{M}}{\partial y}=2 y, \frac{\partial \mathrm{N}}{\partial x}=0\)
Since \(\frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\), (1) is not an exact equation.
Also \(\frac{1}{\mathrm{~N}}\left(\frac{\partial \mathrm{M}}{\partial y}-\frac{\partial \mathrm{N}}{\partial x}\right)=\frac{1}{2 y}(2 y-0)=\frac{1}{2 y}(2 y)=1\) = real number
∴ I.F \(=\exp \left[\int f(x) d x\right]=\exp \left(\int 1 d x\right)=e^x\)
Multiplying(1)by \(e^x \Rightarrow\left(x^2+y^2+2 x\right) e^x d x+2 y e^x d y=0\) …………………(2)
Now (2) is an exact equation (verify) where \(\mathrm{M}_1=\left(x^2+y^2+2 x\right) e^x \text { and } \mathrm{N}_1=2 y e^x\)
∴ G.S. of (2) is \(\int^x\left(x^2+y^2+2 x\right) e^x d x+\int 0 d y=c\) (∵ no term in \(\mathrm{N}_1\) not containing x )
⇒ \(\int x^2 e^x d x+y^2 \int e^x d x+2 \int x e^x d x=c\)
⇒ \(x^2 e^x-\int 2 x e^x d x+y^2 e^x+\int 2 x e^x d x=c \Rightarrow\left(x^2+y^2\right) e^x=c\)
Note. (2) can also be solved by the rearrangement of terms.
(2) \(\Rightarrow\left(x^2+y^2\right) e^x d x+2 e^x(x d x+y d y)=0\)
⇒ \(d\left[\left(x^2+y^2\right) e^x\right]=0 \Rightarrow \int d\left[\left(x^2+y^2\right) e^x\right]=c \Rightarrow\left(x^2+y^2\right) e^x=c\)
Homogeneous Equations Solved Problems Exercise 2.4
Example.3. Solve \(2 x y d y-\left(x^2+y^2+1\right) d x=0\)
Solution.
Given equation is \(2 x y d y-\left(x^2+y^2+1\right) d x=0\) …………………….(1)
where \(\mathrm{M}=-\left(x^2+y^2+1\right) \text { and } \mathrm{N}=2 x y\)
⇒ \(\frac{\partial \mathrm{M}}{\partial y}=-2 y, \frac{\partial \mathrm{N}}{\partial x}\)
= \(2 y \Rightarrow \frac{\partial \mathrm{M}}{\partial y} \neq \frac{\partial \mathrm{N}}{\partial x}\)=> (1) is not an exact equation
Also \(\frac{1}{\mathrm{~N}}\left(\frac{\partial \mathrm{M}}{\partial y}-\frac{\partial \mathrm{N}}{\partial x}\right)=\frac{1}{2 x y}(-2 y-2 y)=\frac{1}{2 x y}(-4 y)=-\frac{2}{x}=f(x)\)
∴ I.F \(=\exp \left[\int f(x) d x\right]=\exp \left[\int \frac{-2}{x} d x\right]=e^{-2 \log x}=e^{\log x^{-2}}=\frac{1}{x^2}\)
Multiplying (1) with \(\frac{1}{x^2} \Rightarrow 2 \frac{y}{x} d y-\left(1+\frac{y^2}{x^2}+\frac{1}{x^2}\right) d x=0\) ……………………(2)
Now (2) is an exact equation where \(\mathrm{M}_1=-\left(1+\frac{y^2}{x^2}+\frac{1}{x^2}\right), \mathrm{N}_1=\frac{2 y}{x}\)
∴ The general solution of (1) is \(\int^x-\left(1+\frac{y^2}{x^2}+\frac{1}{x^2}\right) d x+\int 0 d y=c\) (∵ no terms in \(\mathrm{N}_1\) free from x)
⇒ \(\int-d x+y^2 \int\left(-\frac{1}{x^2}\right) d x+\int\left(-\frac{1}{x^2}\right) d x\) = c
⇒ \(-x+y^2\left(\frac{1}{x}\right)+\frac{1}{x}=c \Rightarrow y^2-x^2+1=c x\)