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

`(\mathrm dy)/(\mathrm dx)=y(y-2)e^x\implies(\mathrm dy)/(y(y-2))=e^x\,\mathrm dx`

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

`\frac1{y(y-2)}=\frac12\left(\frac1{y-2}-\frac1y\right)`

Then

`\displaystyle\int(\mathrm dy)/(y(y-2))=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C`

`\implies\frac12\ln\left|\frac{y-2}y\right|=e^x+C`

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

`\frac12\ln\left|\frac{1-2}1\right|=e^0+C\implies C=-1`

so that the particular solution is

`\frac12\ln\left|\frac{y-2}y\right|=e^x-1\implies\boxed{y=\frac2{1-e^(2e^x-2)}}`