solve the differential equation - $ y \frac{dy}{dx} -(1+y^2)x^2 = 0$

Question

$ y \frac{dy}{dx} -(1+y^2)x^2 = 0$

$\int \frac{y}{1+y^2} dy = \int x^2 dx $ I used separable equations.

$\int \frac{2y}{1+y^2} dy = \frac{x^3}{3} + C$

$\ln | 1 + y^2 | = \frac{x^3}{3} + C $

$y^2 = e^{\frac{x^3}{3} + C} - 1$

Where have I got wrong ?

It should be $e^{\frac{2x^3}{3} + C} -1$

Answer 1

You forgot to multiply by $2$ on the other side when you integrate

$$ \int \frac{2y}{1+y^2} dy = \int 2x^2 dx $$ $$ \ln|1+y^2| = \frac{2x^3}{3} + C $$

Answer 2

I think you have forgotten that $$\int\frac{y}{1+y^2}dy=\frac{1}{2}\int\frac{2y}{1+y^2}dy$$

Answer 3

When you multiply 2, you need to divide by 2 too, to balance the changes you made.

$$ \int \dfrac{y}{1+y^{2}} dy = \dfrac{1}{2}\int \dfrac{2y}{1+y^{2}} dy$$