Question

If y(x) satisfies equations $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ and $y(0)=0,$ then y(1) is

• A. $\frac{2}{3}$
• B. 1
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

Answer: (A)

We have,
$\left(1+x^2\right)\frac{dy}{dx}+2xy-4x^2=0$
$\Rightarrow \frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{4x^2}{1+x^2}$
This is linear differential equation
$\therefore I.F=e^{\int \frac{2x}{1+x^2}dx}$
$=e^{log\left(1+x^2\right)}=1+x^2$
Solution of given differential equation is
$y\left(1+x^2\right)=\int 4x^{2}dx$
$y\left(1+x^2\right)=\frac{4x^3}{3}+C$
put $x=0,y=0$, we get
$C=0$
$\therefore y\left(1+x^2\right)=\frac{4x^3}{3}$
put $x=1,y=\frac{2}{3}$
$\therefore y\left(1\right)=\frac{2}{3}$