Question

Solve:
$y+\frac{d}{dx}(x.y)=x(\sin x+\log x)$

Answer 1

$y+\frac{d}{dx}(x.y)=x(sinx+logx)$

$\Rightarrow y+y+x\frac{dy}{dx}=x(sinx+logx)$

$\Rightarrow 2y+x\frac{dy}{dx}=x(sinx+logx)$

$\Rightarrow \left(\frac{2}{x}\right)y+\frac{dy}{dx}=sinx+logx$

$\Rightarrow \frac{dy}{dx}+\left(\frac{2}{\lambda }\right)y=sinx+logx$

For a differential equation of the first order

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

the solution is

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

Here we have,

$P\left(x\right)=\frac{2}{x}$ and $Q\left(x\right)=sinx+logx$

is the solution of :

$\frac{dy}{dx}+\left(\frac{2}{x}\right)y=sinx+logx$

is given by

${ye}^{\int \frac{2}{x}dx}=\int e^{\int \frac{2}{x}dx}(sinx+logx)dx$

$\Rightarrow {ye}^{ln{x}^2}=\int e^{ln{x}^2}(sinx+logx)dx$

$\Rightarrow x^{2}y=\int x^2(sinx+logx)dx$

$\Rightarrow x^{2}y=\int x^{2}sinxdx+\int x^{2}logxdx$

Now, let's solve the $RHS$ separately,

$\int x^{2}sinxdx$

$=(\int sinxdx)x^2-\int (\int sinxdx\frac{d}{dx}\left(x^2\right))dx$

$=-x^{2}cosx+\int cosx\left(2x\right)dx$

$=-x^{2}cosx+2\int xcosxdx$

$=-x^{2}cosx+2xsinx+2cosx$

Similarly,

$\int x^{2}logxdx$

$=\frac{x^3}{3}logx-\int \frac{x^3}{3}.\frac{1}{x}dx$

$=\frac{x^3}{3}logx-\frac{x^3}{9}$

$\therefore$ the solution is

$x^{2}y={-x}^{2}cosx+2xsinx+2cosx+\frac{x^3}{3}logx-\frac{x^3}{9}+C$