Answer: Either

The fifth roots of unity are the solutions of \(z^5=1\), so

\(z=\cos \frac{2k\pi}{5}+\mathrm{i}\sin \frac{2k\pi}{5}\), where \(k=0,\pm1,\pm2\).

Thus the fifth roots are \(1\), \(\cos \frac{2\pi}{5}\pm \mathrm{i}\sin \frac{2\pi}{5}\), and \(\cos \frac{4\pi}{5}\pm \mathrm{i}\sin \frac{4\pi}{5}\).

Now

\(\left(x-\left[\cos \frac{2\pi}{5}+\mathrm{i}\sin \frac{2\pi}{5}\right]\right)\left(x-\left[\cos \frac{2\pi}{5}-\mathrm{i}\sin \frac{2\pi}{5}\right]\right)=x^2-2\cos \frac{2\pi}{5}\,x+1\).

So the real factors of \(x^5-1\) are

\(\left(x-1\right)\left(x^2-2\cos \frac{2\pi}{5}\,x+1\right)\left(x^2-2\cos \frac{4\pi}{5}\,x+1\right)\).

For \(x^6-x^3+1=0\), let \(u=x^3\). Then

\(u^2-u+1=0\), so \(u=\frac12\pm \mathrm{i}\frac{\sqrt3}{2}=\cos \frac{\pi}{3}\pm \mathrm{i}\sin \frac{\pi}{3}\).

Hence the six roots are

\(x=\cos \frac{\pi}{9}\pm \mathrm{i}\sin \frac{\pi}{9},\; \cos \frac{7\pi}{9}\pm \mathrm{i}\sin \frac{7\pi}{9},\; \cos \frac{13\pi}{9}\pm \mathrm{i}\sin \frac{13\pi}{9}\).

The real factors of \(x^6-x^3+1\) are

\(\left(x^2-2\cos \frac{\pi}{9}\,x+1\right)\left(x^2-2\cos \frac{7\pi}{9}\,x+1\right)\left(x^2-2\cos \frac{13\pi}{9}\,x+1\right)\).

Or

Given \(v=y^3\), the particular solution is \(y^3=5\mathrm{e}^{3x}+3x\mathrm{e}^{3x}+3\mathrm{e}^{-2x}\), so

\(y=\sqrt[3]{5\mathrm{e}^{3x}+3x\mathrm{e}^{3x}+3\mathrm{e}^{-2x}}\).

With \(x=0\), this gives \(y^3=8\), so \(y=2\), as required.

Either

The fifth roots of unity satisfy \(z^5=1\). Writing \(1=\cos 2k\pi+\mathrm{i}\sin 2k\pi\), the roots are

\(z=\cos \frac{2k\pi}{5}+\mathrm{i}\sin \frac{2k\pi}{5},\quad k=0,\pm1,\pm2.\)

So the five roots are

\(1,\ \cos \frac{2\pi}{5}\pm \mathrm{i}\sin \frac{2\pi}{5},\ \cos \frac{4\pi}{5}\pm \mathrm{i}\sin \frac{4\pi}{5}.\)

Now multiply the conjugate pair:

\(\left(x-\left[\cos \frac{2\pi}{5}+\mathrm{i}\sin \frac{2\pi}{5}\right]\right)\left(x-\left[\cos \frac{2\pi}{5}-\mathrm{i}\sin \frac{2\pi}{5}\right]\right)\)

\(=x^2-\left(\cos \frac{2\pi}{5}+\mathrm{i}\sin \frac{2\pi}{5}+\cos \frac{2\pi}{5}-\mathrm{i}\sin \frac{2\pi}{5}\right)x+\left(\cos^2 \frac{2\pi}{5}+\sin^2 \frac{2\pi}{5}\right)\)

\(=x^2-2\cos \frac{2\pi}{5}\,x+1.\)

Similarly, the other conjugate pair gives \(x^2-2\cos \frac{4\pi}{5}\,x+1\). Hence

\(x^5-1=(x-1)\left(x^2-2\cos \frac{2\pi}{5}\,x+1\right)\left(x^2-2\cos \frac{4\pi}{5}\,x+1\right).\)

For \(x^6-x^3+1=0\), let \(u=x^3\). Then

\(u^2-u+1=0.\)

So

\(u=\frac{1\pm \mathrm{i}\sqrt3}{2}=\cos \frac{\pi}{3}\pm \mathrm{i}\sin \frac{\pi}{3}.\)

Since \(x^3=u\), the three cube roots of \(\cos \frac{\pi}{3}+\mathrm{i}\sin \frac{\pi}{3}\) are

\(x=\cos \frac{\pi}{9}+\mathrm{i}\sin \frac{\pi}{9},\ \cos \frac{7\pi}{9}+\mathrm{i}\sin \frac{7\pi}{9},\ \cos \frac{13\pi}{9}+\mathrm{i}\sin \frac{13\pi}{9}.\)

The conjugates are obtained by changing the sign of the sine terms, so the six roots are

\(\cos \frac{\pi}{9}\pm \mathrm{i}\sin \frac{\pi}{9},\quad \cos \frac{7\pi}{9}\pm \mathrm{i}\sin \frac{7\pi}{9},\quad \cos \frac{13\pi}{9}\pm \mathrm{i}\sin \frac{13\pi}{9}.\)

Therefore the real factors are

\(x^6-x^3+1=\left(x^2-2\cos \frac{\pi}{9}\,x+1\right)\left(x^2-2\cos \frac{7\pi}{9}\,x+1\right)\left(x^2-2\cos \frac{13\pi}{9}\,x+1\right).\)

Or

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

\(v'=3y^2\frac{\mathrm{d}y}{\mathrm{d}x}\)

and

\(v''=6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+3y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}.\)

Now the given equation is

\(y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}-6y^2\frac{\mathrm{d}y}{\mathrm{d}x}+2y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+3y^3=25\mathrm{e}^{-2x}.\)

Multiply by 3:

\(3y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}-18y^2\frac{\mathrm{d}y}{\mathrm{d}x}+6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+9y^3=75\mathrm{e}^{-2x}.\)

Using the expressions for \(v'\) and \(v''\), this becomes

\(v''-6v'+9v=75\mathrm{e}^{-2x}.\)

Now solve the linear equation.

The complementary function comes from

\(m^2-6m+9=0\), so \((m-3)^2=0\) and

\(v_c=A\mathrm{e}^{3x}+Bx\mathrm{e}^{3x}.\)

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

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

Substitute into \(v''-6v'+9v=75\mathrm{e}^{-2x}\):

\(4k+12k+9k=75\), so \(25k=75\) and \(k=3\).

Hence

\(v=A\mathrm{e}^{3x}+Bx\mathrm{e}^{3x}+3\mathrm{e}^{-2x}.\)

Since \(v=y^3\), we have

\(y^3=A\mathrm{e}^{3x}+Bx\mathrm{e}^{3x}+3\mathrm{e}^{-2x}.\)

Use \(x=0, y=2\). Then \(v=8\), so

\(8=A+3\), giving \(A=5\).

Differentiate \(v\):

\(v'=3A\mathrm{e}^{3x}+B\mathrm{e}^{3x}+3Bx\mathrm{e}^{3x}-6\mathrm{e}^{-2x}.\)

At \(x=0\), \(y=2\) and \(\frac{\mathrm{d}y}{\mathrm{d}x}=1\), so

\(v'=3y^2y'=3\cdot 2^2\cdot 1=12.\)

Also, with \(A=5\), at \(x=0\):

\(12=15+B-6\), so \(B=3\).

Therefore

\(y^3=5\mathrm{e}^{3x}+3x\mathrm{e}^{3x}+3\mathrm{e}^{-2x},\)

and since \(y=2\) when \(x=0\), the required particular solution is

\(y=\sqrt[3]{5\mathrm{e}^{3x}+3x\mathrm{e}^{3x}+3\mathrm{e}^{-2x}}.\)