Finding a y(x) that satisfies $ y(x) = \int_0^x \! \left(\frac{t} {y(t)+1}\right)^2 \, \mathrm{d}t $

Question

I'm having problem with finding a y(x) that satisfies $$ y(x) = \int_0^x \! \left(\frac{t} {y(t)+1}\right)^2 \, \mathrm{d}t $$ Here is what I tried to do. $$ y(x) = \int_ \! \left(\frac{x} {y(x)+1}\right)^2 \, \mathrm{d}x $$ $$ y'(x) = \left(\frac{x} {y(x)+1}\right)^2 $$ $$ \frac {y(x)^3} {3} + y(x)^2 + y(x) = \frac {x^3} {3} $$ Here I can't figure out how to move forward. Anyone who can help me with solving the question or at least giving me a hint on what kind of method I should use to solve this question?

Answer 1

Going from $y'(x) = \left(\frac{x} {y(x)+1}\right)^2$, you can rewrite this as

$$(y(x) + 1)^2 \frac{\mathrm{d}y}{\mathrm{d}x} = x^2,$$

so that integrating both sides with respect to $x$ gives

$$\int (y+1)^2\mathrm{d}y = \int x^2 \mathrm{d}x.$$

In other words, you were almost there, but it's not always best to distribute out everything.