Solve initial value problem $u^{'}(t)=\frac{2(u(t))^{2}}{3(t+1)^{2}}$, $\ u(1)=3$

Question

So I just proved that every initial value problem of the form

$\hspace{2cm}y^{'}(t)=f(t)\hspace{3cm} y(0)=y_{0}\hspace{1cm}(1.1)$

for $f\in C^{0}(I,\mathbb{R})$ has a unique solution on $I$. It is the antiderivative $F$ of $f$.

Now I'm confronted with the following problem:

Transform the initial value problem

$\hspace{2cm} u^{'}(t)=\dfrac{2(u(t))^{2}}{3(t+1)^{2}}\hspace{3cm} u(1)=3$

in one of the form $(1.1)$ and determine the solution $u(t)$ for $t\geq 1$.

I'm not quite sure how to advance here. It seems to me rather like a lot of guess work. Or is there a smart way to do it. It should be noted that I just started learning about ordinary differential equations so we haven't proved anything yet, just some basic definitions.

Answer 1

Hint:

Equations of the form $y'(t)=\dfrac{dy}{dt}=f(y)\,g(t)$ are called separable.

If you rewrite as

$$\frac{dy}{f(y)}=g(t)\,dt,$$ the solution should seem more obvious.

Answer 2

Write your equation in the form $$\frac{u'(t)}{\frac{u(t)^2}{3}}=\frac{2}{(1+t)^2}$$