Question

Solve the following differential equation:
$x\cos ydy=\left({xe}^x\log x+e^x\right)dx$

Answer 1

$xcosydy=(x{e}^{x}logx+e^x)dx$

$\int cosydy=\int \frac{x{e}^{x}logx+e^x}{x}dx$

$=\int e^{x}logxdx+\int \frac{e^{x}dx}{x}$

$=\int e^{x}logxdx$

$=e^{x}logx=\int \frac{e^x}{x}dx$

$siny=e^{x}logx-\int \frac{e^x}{x}dx+\int \frac{e^x}{x}dx$

$\therefore e^{x}logx=siny+c$