Question

The solution of the differential equation $\frac{dy}{dx}+\frac{2yx}{1+x^2}=\frac{1}{(1+x^2{)}^2}$ is.

• A. $y(1+x^2)=C+ta{n}^{-1}x$
• B. $\frac{y}{(1+x^2)}=C+ta{n}^{-1}x$
• C. $ylog(1+x^2)=C+si{n}^{-1}x$
• D. $y(1+x^2)=C+si{n}^{-1}x$

Answer 1

The correct option is A $y(1+x^2)=C+ta{n}^{-1}x$

$\frac{dy}{dx}+\frac{2yx}{1+x^2}=\frac{1}{(1+x^2{)}^2}$

which is a linear differential equation.

Here, $P=\frac{2x}{1+x^2},Q=\frac{1}{(1+x^2{)}^2}$

Now, $IF=e^{\int Pdx}$

$=e^{\int \frac{2x}{1+x^2}dx}=e^{log(1+x^2)}=(1+x^2)$

$\therefore$ Solution of differential equation is
$y.(1+x^2)=\int \frac{1}{(1+x^2{)}^2}.(1+x^2)dx+C$

$⟹y(1+x^2)=\int \frac{1}{1+x^2}dx+C$

$⟹y(1+x^2)=ta{n}^{-1}x+C$

Hence, option A is correct.