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

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

Differentiating again,

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

Now start with

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

Rewrite the left-hand side as

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

Using \(v''=2\dfrac{dy}{dx}+x\dfrac{d^2y}{dx^2}\), \(v'=y+x\dfrac{dy}{dx}\) and \(v=xy\), this becomes

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

To solve this, first find the complementary function from

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

So

\((m+3)(m-1)=0\), giving \(m=-3\) and \(m=1\).

Hence the complementary function is

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

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

Then

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

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

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

So

\(5ke^{2x}=10e^{2x}\), hence \(k=2\).

Therefore

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

Since \(v=xy\),

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