Question

The solution of $x\frac{dy}{dx}+ylogy=xy{e}^x$ is

• A. $logy=(x-1)e^x+c$
• B. $(x-1)logy=x{e}^x+c$
• C. $xlogy=(x+1)e^x+c$
• D. $xlogy=(x-1)e^x+c$

Answer 1

The correct option is D $xlogy=(x-1)e^x+c$

$x\frac{dy}{dx}+ylny=xy{e}^x$

$\frac{x}{y}\frac{dy}{dx}+lny=x{e}^x$

Let $xlny=t$

$lny+\frac{x}{y}\frac{dy}{dx}=\frac{dt}{dx}$

$\frac{dt}{dx}=x.e^x$

$\int dt=\int x-e^x$

$t=x{e}^x-\int e^{x}dx$

$t=e^x(x-1)+c$

prove value of 1

$xlny=e^x(x-1)+c$