Differential Equations (how to proceed) $$y^2+\frac{dy}{dx}=4$$ $y=0$, when $x=\ln 2$
$$\frac{dy}{dx}=-y^2+4$$
$$\frac{\frac{dy}{dx}}{-y^2+4}=1$$
$$\int \frac{\frac{dy}{dx}}{-y^2+4}dx=\int 1dx$$
$$-\frac{1}{4}\ln (-y+2)+\frac{1}{4}\ln (y+2)=x+c_1$$
How to proceed to solve y?

$$ \begin{align}x - \ln 2 = \int_{\ln 2}^x dx &= \int_0^y\frac{dy}{4-y^2}\\& = \frac 14\int_0^y \left(\frac{1}{2-y} + \frac 1{2+y}\right)\\&=\frac 14 \left(\ln(2+y) - \ln(2-y)\right)\\&=\frac 14\ln\left(\frac{2+y}{2-y}\right)\end{align}$$
that is $$\frac{2 + y}{2-y}=\frac{e^{4x}}{16}\to \frac{y}{2} = \frac{e^{4x}-16}{e^{4x} + 16} $$ finally we have $$ y = \frac{2(e^{4x}-16)}{e^{4x} + 16}$$
