What is the solution for y^{''}-3y^'+2y=4e^x ?

The solution for y^{''}-3y^'+2y=4e^x is y=c_{1}e^{2x}+c_{2}e^x-4xe^x

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} - 4 x e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} - 4 x e^{x}

y = c_{1} e^{2 x} + c_{2} e^{x} - 4 x e^{x}

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

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 3 y = cos (2 x)

y^{^{''}} + 4 y = 4, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 6, y^{^{'}} (0) = 8

y^{^{''}} + 2 y^{^{'}} + 5 y = 1 - t, y (0) = 0, y^{^{'}} (0) = 2

What is the solution for y^{''}-3y^'+2y=4e^x ?
The solution for y^{''}-3y^'+2y=4e^x is y=c_{1}e^{2x}+c_{2}e^x-4xe^x