Question

Find the I.F of $x\frac{dy}{dx}+y(1+x)=1$

• A. $x.e^x$
• B. $e^x/x$
• C. $x+logx$
• D. $xlogx$

Answer 1

Answer: (A)

$x.e^x$

Given differential equation is $x\frac{dy}{dx}+y(1+x)=1$

Dividing both sides of above D.E. by $x$ we get
$\frac{dy}{dx}+\frac{y(1+x)}{x}=\frac{1}{x}$

This is in the form of linear D.E. i.e $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$, where

$P\left(x\right)=\frac{1+x}{x}=1+\frac{1}{x}$ and $Q\left(x\right)=\frac{1}{x}$

and $\text{I.F.}=e^{\int P\left(x\right)dx}$

$⟹\text{I.F.}=e^{\int 1+\frac{1}{x}dx}$

$=e^{x+\log x}$

$=e^x{e}^{\log x}=e^{x}x$

$⟹\text{I.F.}=x{e}^x$

Hence, option A is correct.