Answer: The general solution is

\(y=\dfrac{1}{x}\left(Ae^{-3x}+Be^x+2e^{2x}\right),\)

where \(A\) and \(B\) are arbitrary constants.

Let \(v=xy\). Then

\(\dfrac{dv}{dx}=y+x\dfrac{dy}{dx}\)

and differentiating again,

\(\dfrac{d^2v}{dx^2}=2\dfrac{dy}{dx}+x\dfrac{d^2y}{dx^2}.\)

Using the given equation

\(x\dfrac{d^2y}{dx^2}+(2x+2)\dfrac{dy}{dx}+(2-3x)y=10e^{2x},\)

rewrite the left-hand side as

\(\left(x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}\right)+2\left(y+x\dfrac{dy}{dx}\right)-3xy.\)

Since

\(x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}=\dfrac{d^2v}{dx^2}, \qquad y+x\dfrac{dy}{dx}=\dfrac{dv}{dx}, \qquad xy=v,\)

this becomes

\(\dfrac{d^2v}{dx^2}+2\dfrac{dv}{dx}-3v=10e^{2x},\)

as required.

Now solve

\(v''+2v'-3v=10e^{2x}.\)

For the complementary function, the auxiliary equation is

\(m^2+2m-3=0,\)

so

\((m+3)(m-1)=0,\)

giving \(m=-3\) and \(m=1\). Hence

\(v_c=Ae^{-3x}+Be^x.\)

For a particular integral, try \(v_p=ke^{2x}\). Then

\(v_p'=2ke^{2x}, \qquad v_p''=4ke^{2x}.\)

Substituting gives

\(4ke^{2x}+2(2ke^{2x})-3(ke^{2x})=10e^{2x},\)

so

\((4k+4k-3k)e^{2x}=10e^{2x}\Rightarrow 5k=10\Rightarrow k=2.\)

Therefore

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

Since \(v=xy\),

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