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

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

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} - 3 x e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} - 3 x e^{x}

y = c_{1} e^{2 x} + c_{2} e^{x} - 3 x e^{x}

y^{^{''}} + 36 y = 5 sec (6 x)

y^{^{''}} - 4 y^{^{'}} = (t^{2} + 1) sin (2 t)

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t)

y^{^{''}} - 2 y^{^{'}} + y = (x + 1) e^{x}

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

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