What is the solution for y^{''}-3y^'+2y=2xe^x ?

The solution for y^{''}-3y^'+2y=2xe^x is y=c_{1}e^{2x}+c_{2}e^x-e^xx^2-2e^xx

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y^{''}-3y^'+2y=2xe^x

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 x e^{x}

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x^{2} - 2 e^{x} x

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = 2 x e^{x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x^{2} - 2 e^{x} x

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x^{2} - 2 e^{x} x

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{t}, y (0) = 2, y^{^{'}} (0) = 4

y^{^{''}} - y = 2 t - 4

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} + 8 y = e^{- 3 x} + 8 x^{2}

\frac{1}{6} y^{^{''}} + 6 y = tan (6 t) - \frac{1}{3} e^{3 t}

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- 2 x} sec (x)

What is the solution for y^{''}-3y^'+2y=2xe^x ?
The solution for y^{''}-3y^'+2y=2xe^x is y=c_{1}e^{2x}+c_{2}e^x-e^xx^2-2e^xx