Question

Find the general solution of the differential equation $x\frac{dy}{dx}+y-x+xy\cot x=0(x\ne 0)$

Answer 1

Given differential equation :

$x\frac{dy}{dx}+y-x+xy\cot x=0$

$\frac{dy}{dx}+\frac{y}{x}-1+y\cot x=0$

$\frac{dy}{dx}+y(\frac{1}{x}+\cot x)-1=0$

$\frac{dy}{dx}+(\frac{1}{x}+\cot x)y=1$

Given differential equation is of the form
$\frac{dy}{dx}+Py=Q$

By comparing both the equations, we get

$P=(\frac{1}{x}+\cot x)$ and $Q=1$

The general solution of the given differential equation is

$y(I.F)=\int (Q\times I.F.)dx+c$ ....(i)

Firstly, we need to find I.F.

$I.F.=e^{\int pdx}$

$I.F.=e^{\int (\frac{1}{x}+\cot x)}dx$

$I.F.=e^{\text{log}x+\text{log}\sin x}$

$I.F.=e^{\text{log}\left(x\sin x\right)}(\because \text{log}a+\text{log}b=\text{log}ab)$

$I.F.=x\sin x(\because e^{\text{log}x}=x)$

Substituting the value of I.F in (i), we get

$y.x\sin x=\int 1.x\sin xdx+c$

Using Integration by parts :
$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-f\left[f^{\prime }\right(x)\int g(x\left)dx\right]dx$

Taking $f\left(x\right)=x$ and $g\left(x\right)=\sin x$

$yx\sin x=x\int \sin xdx-\int [1.\int \sin x.dx]dx+c$

$yx\sin x=x(-\cos x)-\int 1.(-\cos x)dx+c$

$yx\sin x=-x\cos x+\int \cos xdx+c$

$yx\sin x=-x\cos x+\sin x+c$

$y=\frac{-x\cos x}{x\sin x}+\frac{\sin x}{x\sin x}+\frac{c}{x\sin x}$

$y=-\cot x+\frac{1}{x}+\frac{c}{x\sin x}$
Hence, the required general solution is
$y=-\cot x+\frac{1}{x}+\frac{c}{x\sin x}$


$\therefore y=(1+x^2{)}^{-1}\text{log}|\sin x|+c(1+x^2{)}^{-1}$