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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} = cos (x)

y^{^{''}} + 4 y = 2

y^{^{''}} + 36 y = 24 sin (6 t)

y^{^{'''}} - 9 y^{^{'}} = t e^{- 3 t} + 4 cos (3 t)

0.05 y^{^{''}} - 0.3 y^{^{'}} + 0.4 y = 11

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