Question

Let $y=f\left(x\right)$ defined on $R$ satisfy the differential equation $(1+x^2)\frac{dy}{dx}+2xy=2x$ and $y\left(0\right)=2.$ Then which of the following statements is/are correct?

• A. $f\left(x\right)$ is neither even nor odd function.
• B. $f\left(x\right)$ increases on $(-\infty ,0)$ and decreases on $(0,\infty )$.
• C. The $x-$intercept of normal on the graph of $y=f\left(x\right)$ at $x=1$ equals $\frac{1}{4}$.
• D. The area bounded by $y=f\left(x\right)$ with $x-$axis between the ordinates $x=0$ and $x=1$ equals $\left(\frac{\pi +4}{4}\right)$ sq. units.

Answer 1

Answer: (B), (C), (D)

The area bounded by $y=f\left(x\right)$ with $x-$axis between the ordinates $x=0$ and $x=1$ equals $\left(\frac{\pi +4}{4}\right)$ sq. units.

$(1+x^2)\frac{dy}{dx}+2xy=2x$
$\Rightarrow \frac{dy}{dx}+\left(\frac{2x}{1+x^2}\right)y=\frac{2x}{1+x^2}$
which is a linear differential equation.
$\int \frac{2x}{1+x^2}dx=\ln (1+x^2)$
$\therefore I.F.=e^{\ln (1+x^2)}=1+x^2$
Now, the general solution is
$y(1+x^2)=\int \left(\frac{2x}{1+x^2}\right).(1+x^2)dx+C$
$\Rightarrow y(1+x^2)=x^2+C$
$y\left(0\right)=2\Rightarrow 2=0+C$
$\Rightarrow C=2$
$\therefore f\left(x\right)=\frac{x^2+2}{x^2+1}$


Since $f(-x)=f\left(x\right)$
$\therefore f\left(x\right)$ is an even function.


$f\left(x\right)=\frac{x^2+2}{x^2+1}=1+\frac{1}{x^2+1}$
$f^{\prime }\left(x\right)=\frac{-2x}{(x^2+1{)}^2}$
$\therefore f^{\prime }\left(x\right)>0\forall x<0$ and $f^{\prime }\left(x\right)<0\forall x>0$
Hence, $f\left(x\right)$ increases on $(-\infty ,0)$ and decreases on $(0,\infty )$.


$f^{\prime }\left(x\right)=\frac{-2x}{(x^2+1{)}^2}$
$f^{\prime }\left(1\right)=-\frac{1}{2}$
$\therefore$ Slope of normal $=2$
Equation of normal at $(1,\frac{3}{2})$ is
$y-\frac{3}{2}=2(x-1)$
For $x-$intercept, put $y=0$
$-\frac{3}{2}=2(x-1)$
$\Rightarrow x=\frac{1}{4}$


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


Area $=\underset{0}{\overset{1}{\int }}f\left(x\right)dx$
$=\underset{0}{\overset{1}{\int }}(1+\frac{1}{1+x^2})dx$
$=[x+{\tan }^{-1}x{]}_0^1$
$=\left(\frac{\pi +4}{4}\right)$ sq. units