Answer: (a) \(\displaystyle \frac{\mathrm{d}^2v}{\mathrm{d}x^2}+\frac{\mathrm{d}v}{\mathrm{d}x}+4v=4\mathrm{e}^{-2x}\).

(b) \(\displaystyle y=\left(\frac13\mathrm{e}^{-x/2}\cos\left(\frac{\sqrt{15}}{2}x\right)+\frac23\mathrm{e}^{-2x}\right)^{1/4}\).

(a) Since \(v=y^4\),

\(\displaystyle \frac{\mathrm{d}v}{\mathrm{d}x}=4y^3\frac{\mathrm{d}y}{\mathrm{d}x}.\)

Differentiating again,

\(\displaystyle \frac{\mathrm{d}^2v}{\mathrm{d}x^2}=4y^3\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+12y^2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2.\)

Hence

\(\displaystyle \frac{\mathrm{d}^2v}{\mathrm{d}x^2}+\frac{\mathrm{d}v}{\mathrm{d}x}+4v\)

\(\displaystyle =4y^3\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+12y^2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+4y^3\frac{\mathrm{d}y}{\mathrm{d}x}+4y^4\)

\(\displaystyle =4\left(y^3\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+3y^2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+y^3\frac{\mathrm{d}y}{\mathrm{d}x}+y^4\right).\)

Using the given equation, this becomes

\(\displaystyle \frac{\mathrm{d}^2v}{\mathrm{d}x^2}+\frac{\mathrm{d}v}{\mathrm{d}x}+4v=4\mathrm{e}^{-2x}.\)

(b) We solve

\(\displaystyle v''+v'+4v=4\mathrm{e}^{-2x}.\)

The auxiliary equation is

\(\displaystyle m^2+m+4=0,\)

so the complementary solution is

\(\displaystyle v_c=\mathrm{e}^{-x/2}\left(A\cos\left(\frac{\sqrt{15}}{2}x\right)+B\sin\left(\frac{\sqrt{15}}{2}x\right)\right).\)

For a particular solution, try \(v_p=k\mathrm{e}^{-2x}\). Then

\(\displaystyle v_p'=-2k\mathrm{e}^{-2x},\qquad v_p''=4k\mathrm{e}^{-2x}.\)

Substituting,

\(\displaystyle 4k\mathrm{e}^{-2x}-2k\mathrm{e}^{-2x}+4k\mathrm{e}^{-2x}=4\mathrm{e}^{-2x},\)

so \(6k=4\) and therefore

\(\displaystyle k=\frac23.\)

Thus

\(\displaystyle v=\mathrm{e}^{-x/2}\left(A\cos\left(\frac{\sqrt{15}}{2}x\right)+B\sin\left(\frac{\sqrt{15}}{2}x\right)\right)+\frac23\mathrm{e}^{-2x}.\)

Since \(v=y^4\), when \(x=0\) and \(y=1\), we have \(v=1\). Hence

\(\displaystyle 1=A+\frac23,\)

so

\(\displaystyle A=\frac13.\)

Now

\(\displaystyle v'=\mathrm{e}^{-x/2}\left(-\frac{\sqrt{15}}{2}A\sin\left(\frac{\sqrt{15}}{2}x\right)+\frac{\sqrt{15}}{2}B\cos\left(\frac{\sqrt{15}}{2}x\right)\right)-\frac12\mathrm{e}^{-x/2}\left(A\cos\left(\frac{\sqrt{15}}{2}x\right)+B\sin\left(\frac{\sqrt{15}}{2}x\right)\right)-\frac43\mathrm{e}^{-2x}.\)

Also, since \(v=y^4\),

\(\displaystyle v'=4y^3\frac{\mathrm{d}y}{\mathrm{d}x}.\)

At \(x=0\), \(y=1\) and \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac38\), so

\(\displaystyle v'(0)=4(1)^3\left(-\frac38\right)=-\frac32.\)

Substituting \(x=0\) into the expression for \(v'\),

\(\displaystyle -\frac32=\frac{\sqrt{15}}{2}B-\frac12A-\frac43.\)

With \(A=\frac13\),

\(\displaystyle -\frac32=\frac{\sqrt{15}}{2}B-\frac16-\frac43=\frac{\sqrt{15}}{2}B-\frac32,\)

so \(B=0\).

Therefore

\(\displaystyle y^4=\frac13\mathrm{e}^{-x/2}\cos\left(\frac{\sqrt{15}}{2}x\right)+\frac23\mathrm{e}^{-2x}.\)

Since \(y(0)=1\), we take the positive fourth root:

\(\displaystyle y=\left(\frac13\mathrm{e}^{-x/2}\cos\left(\frac{\sqrt{15}}{2}x\right)+\frac23\mathrm{e}^{-2x}\right)^{1/4}.\)