Question

A function $y=f\left(x\right)$ satisfies $(x+1).f^{\prime }\left(x\right)-2(x^2+x)f\left(x\right)=\frac{e^{x^2}}{(x+1)},\forall x>-1$. If $f\left(0\right)=5$, then $f\left(x\right)$ is

• A. $\left(\frac{3x+5}{x+1}\right).e^{x^2}$
• B. $\left(\frac{6x+5}{x+1}\right).e^{x^2}$
• C. $\left(\frac{6x+5}{(x+1{)}^2}\right).e^{x^2}$
• D. $\left(\frac{5+6x}{x+1}\right).e^{x^2}$

Answer 1

The correct option is B $\left(\frac{6x+5}{x+1}\right).e^{x^2}$

Given,

$(x+1).f^{\prime }\left(x\right)-2(x^2+x)f\left(x\right)=\frac{e^{x^2}}{(x+1)}$

or, $f^{\prime }\left(x\right)-2x.f\left(x\right)=\frac{e^{x^2}}{(x+1{)}^2}$

or, $e^{-x^2}{f}^{\prime }\left(x\right)-2x.e^{-x^2}f\left(x\right)=\frac{1}{(x+1{)}^2}$

or, $d\left(e^{-x^2}f\right(x\left)\right)=\frac{dx}{(x+1{)}^2}$

Now integrating both sides we get,

$e^{-x^2}f\left(x\right)=-\frac{1}{1+x}+c$.......(1) [ Where $c$ is integrating constant]

Given $f\left(0\right)=5$

This gives from the solution ,

$c=6$.

Using this value in (1) we get,

$e^{-x^2}f\left(x\right)=\frac{6x+5}{x+1}$

or, $f\left(x\right)=\frac{6x+5}{x+1}{e}^{x^2}$.