Solve x²y''+4xy'+2y=eˣ

x² y'' + 4x y' + 2y = eˣ

Quick method: recognise the left side

Since (x²y)'' = x²y'' + 4xy' + 2y, the equation is

d²(x²y)/dx² = eˣ

Integrate twice:

d(x²y)/dx = eˣ + c₁ x²y = eˣ + c₁x + c₂

y = eˣ/x² + c₁/x + c₂/x²

Check by the standard (Cauchy–Euler) method

Put z = log x, D = d/dz: (D(D − 1) + 4D + 2)y = e^(eᶻ), i.e. (D + 1)(D + 2)y = e^(eᶻ).

CF = c₁e^(−z) + c₂e^(−2z) = c₁/x + c₂/x², which matches the terms above. The particular integral gives eˣ/x², so the general solution is the same.