Question

x $\frac{dy}{dx}$ - 2y = x${}^2$ + sin $\frac{1}{x^2}$ x > 0.

Answer 1

$x\frac{dy}{dx}-2y=x^2+\sin \frac{1}{x^2},x>0$

$\frac{dy}{dx}-\frac{2y}{x}=x+\frac{1}{x}\sin \frac{1}{x^2}$

The above is a linear equation of the form

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

Where, $P\left(x\right)=-\frac{2}{x}$

$Q\left(x\right)=x+\frac{1}{x}\sin \frac{1}{x^2}$

Then the solution becomes

$\frac{dy}{dx}\{y.e^{\int Pdx}\}={Q\left(x\right)e}^{\int Pdx}$

$\Rightarrow y.e^{\int Pdx}=\int Q\left(x\right)e^{\int Pdx}dx$

$\Rightarrow y.e^{\int \frac{-2}{x}dx}=\int (x+\frac{\sin \frac{1}{x^2}}{x})e^{\int \frac{-2}{x}dx}dx$

$\Rightarrow y.e^{-2Inx}=\int (x+\frac{1}{x}\sin \frac{1}{x^2})e^{-2Inx}dx$

$\Rightarrow y.e^{In{x}^{-2}}=\int (x+\frac{1}{x}\sin \frac{1}{x^2})e^{In{x}^{-2}}dx$

$\Rightarrow y.\frac{1}{x^2}=\int (x+\frac{1}{x}\sin \frac{1}{x^2}).\frac{1}{x^2}dx$

$\Rightarrow \frac{y}{x^2}=\int \frac{1}{x}dx+\int \frac{1}{x^3}\sin \frac{1}{x^2}dx$

$\Rightarrow \frac{y}{x^2}=Inx+\int \frac{1}{x^3}\sin \frac{1}{x^2}dx$

Say $\frac{1}{x^2}=t\Rightarrow dt=\frac{-2}{x^3}dx$

$\therefore$ $\Rightarrow \frac{y}{x^2}=Inx-\frac{1}{2}\int sintdt$

$\Rightarrow \frac{y}{x^2}=Inx+\frac{1}{2}\cos \frac{1}{x^2}+c$