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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 25 y = 10 cos (5 x)

y^{^{''}} + 0.01 (20 - y) = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 x + 7

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

\frac{1}{2} y^{^{''}} + \frac{1}{2} y^{^{'}} + 6 y = 10 cos (3 x)

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