What is the solution for y^{''}-5y^'+6y=-5e^{3x} ?

The solution for y^{''}-5y^'+6y=-5e^{3x} is y=c_{1}e^{3x}+c_{2}e^{2x}-5xe^{3x}

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y^{''}-5y^'+6y=-5e^{3x}

y^{^{''}} - 5 y^{^{'}} + 6 y = - 5 e^{3 x}

y = c_{1} e^{3 x} + c_{2} e^{2 x} - 5 x e^{3 x}

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = - 5 e^{3 x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} - 5 x e^{3 x}

y = c_{1} e^{3 x} + c_{2} e^{2 x} - 5 x e^{3 x}

y^{^{''}} - 2 y^{^{'}} + y = t^{2} + 1

y^{^{''}} + 8 y^{^{'}} + 16 y = 6 cos (4 t)

x^{^{''}} + 4 x = - 6 sin (2 t) + 3 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 12 = x

y^{^{''}} + y^{^{'}} = \frac{1}{sin ( x )}

What is the solution for y^{''}-5y^'+6y=-5e^{3x} ?
The solution for y^{''}-5y^'+6y=-5e^{3x} is y=c_{1}e^{3x}+c_{2}e^{2x}-5xe^{3x}