Question

Solve : $x\frac{dy}{dy}+y-x+xycotx=0$

Answer 1

We have,

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

Dividing both side by x and we get,

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

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

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

Comparing that,

$\frac{dy}{dx}+Py=Q$

Then,

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

Now,

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

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

$=e^{\int \frac{1}{x}dx+\int \cot xdx}$

$=e^{\log x+\mathrm{logsin}x}$

$=e^{\log x\sin x}$

$=x\sin x(\therefore e^{\log x}=x)$

Then solution is.

$y\times I.F.=\int Q\times I.F.dx+C$

$y\left(x\sin x\right)=\int x\sin xdx+C$

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

$=x(-\cos x)+\int \left(\cos x\right)dx+C$

$y\times I.F.=-x\cos x+\sin x+C$

Put the value of y

Then,

$y.x\sin x=-x\cos x+\sin x+C$

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

$y=-\cot x+\frac{1}{x}+C_1$

$Where{C}_1=\frac{C}{x\sin x}$

Hence, this is the solution.