Question

If the solution of the differential equation $x\frac{dy}{dx}+y=x{e}^x$ be, $xy=e^x\phi \left(x\right)+c$ then $\phi \left(x\right)$ is equal to:

• A. $x+1$
• B. $x-1$
• C. $1-x$
• D. $x$

Answer 1

Answer: (B)

$x-1$

Given differential equation is, $x\frac{dy}{dx}+y=x{e}^x$

Divide each sides by $x$

$\frac{dy}{dx}+\frac{1}{x}y=e^x$

Comparing with linear differential equation $\frac{dy}{dx}+py=Q$

IF $=e^{\int Pdx}=e^{\int \frac{1}{x}dx}=e^{\ln x}=x$

Hence required solution is,

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

$\Rightarrow xy=x{e}^x-e^x+c$, use by parts to integrate RHS integral

$\Rightarrow xy=e^x(x-1)+c$

Hence required function is $x-1$