Answer: The general solution is \(v=Ae^{-5x}+Be^{3x}-\frac{3}{2}e^{-x}\), so since \(v=y^3\),

\(y=\left(Ae^{-5x}+Be^{3x}-\frac{3}{2}e^{-x}\right)^{1/3}\).

Let \(v=y^3\). Then

\(\frac{dv}{dx}=3y^2\frac{dy}{dx}\)

and differentiating again,

\(\frac{d^2v}{dx^2}=6y\left(\frac{dy}{dx}\right)^2+3y^2\frac{d^2y}{dx^2}\).

Now substitute these into the given differential equation:

\(y^2\frac{d^2y}{dx^2}+2y^2\frac{dy}{dx}+2y\left(\frac{dy}{dx}\right)^2-5y^3=8e^{-x}.\)

Since \(v=y^3\), we have \(v'=3y^2y'\) and \(v''=3y^2y''+6y(y')^2\), so

\(\frac{1}{3}v''+\frac{2}{3}v'-5v=8e^{-x}\)

after the \(y(y')^2\) terms cancel. Multiplying through by 3 gives

\(\frac{d^2v}{dx^2}+2\frac{dv}{dx}-15v=24e^{-x}.\)

To solve this linear differential equation, first consider the complementary function. The auxiliary equation is

\(m^2+2m-15=0\)

which factorises as

\((m+5)(m-3)=0\).

Hence \(m=-5\) or \(m=3\), so

\(v_{\text{CF}}=Ae^{-5x}+Be^{3x}.\)

For a particular integral, try \(v=ke^{-x}\). Then

\(v'=-ke^{-x},\qquad v''=ke^{-x}.\)

Substituting into \(v''+2v'-15v=24e^{-x}\) gives

\(ke^{-x}-2ke^{-x}-15ke^{-x}=24e^{-x},\)

so

\(-16k=24\quad\Rightarrow\quad k=-\frac{3}{2}.\)

Therefore

\(v=Ae^{-5x}+Be^{3x}-\frac{3}{2}e^{-x}.\)

Finally, since \(v=y^3\),

\(y=\left(Ae^{-5x}+Be^{3x}-\frac{3}{2}e^{-x}\right)^{1/3}.\)