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}$