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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 4 y^{^{''}} + y^{^{'}} + 6 y = 8 cos (x) + 18 x + 3

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

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

3 y^{^{''}} + y^{^{'}} - 2 y = x^{2} - 3 x + 7

y^{^{'''}} + y^{^{'}} - 10 y = x e^{2 x}

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