Easiest way to solve D.E.

Question

I was given this question in class today, $$y’’ -\frac{(y’)^2}{y}=y.$$ I got to $$ y\cdot y’’ -(y’)^2 -y^2 =0$$ But I mean I’m lost. I’m sure it’s a quick computation.

Answer 1

$$y’’ -\frac{(y’)^2}{y}=y.$$ $$\dfrac {y’’y -{(y’)^2}}{y^2}=1$$ Then use $$\frac {f'g-fg'}{g^2}=\left (\frac {f}{g} \right )'$$ With $f=y'$ and $g=y$ $$\implies \left (\frac {y'}{y} \right )'=1$$ Integrate.

Answer 2

Hint:

Compute $$\left (\frac {y'}{y}\right)'.$$

Answer 3

Hint:

When writing

$$yy''-y'^2-y^2=0$$

you are not so far, because you can notice that

$$yy''+y'^2$$

is the derivative of $yy'$. Anyway, the minus sign does not match, and you can think of a quotient. Indeed

$$\left(\frac{y'}y\right)'=\frac{yy''-y'^2}{y^2}.$$

The rest is yours.