Solving Initial Value Problems

Question

Considering the following IVP

$y’= \frac{1}{1+x^{2}} -2y^{2} \hspace{2cm} x \in (0,1]$

$y(0)=0$

I need to show that the solution is $ \frac{x}{1+x^{2}}$

Answer 1

Hint: Checking $y(0)=0/(1+0^2)=0$ shows that the initial value is correct. Then plug

$$y(x) = \dfrac{x}{1+x^2}$$ into the differential equaiton and check if the equation is fullfilled.

EDIT: We know that

$$y'(x) = \dfrac{1-x^2}{(1+x^2)^2}.$$ At the same time the right hand side evaluates to

$$ \dfrac{1}{1+x^2}-2y^2=\dfrac{1}{1+x^2}-2\dfrac{x^2}{(1+x^2)^2}=\dfrac{1-x^2}{(1+x^2)^2}.$$

Hence, the left side equals the right side which verifies that $y(x) = x/(1+x^2)$ is a solution to the differential equation.