Question

Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.

Answer 1

This equation is separable, as

$\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx$

Integrate both sides; on the left, expand the fraction as

$\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)$

Then

$\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C$

$\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C$

Since $y(0)=1$, we get

$\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1$

so that the particular solution is

$\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}$