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Irina18 [472]
3 years ago
9

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

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
4 0

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}}}

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