What is the solution for y^{''}-4y^'+4y=12x+7 ?

The solution for y^{''}-4y^'+4y=12x+7 is y=c_{1}e^{2x}+c_{2}xe^{2x}+3x+19/4

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y^{''}-4y^'+4y=12x+7

y^{^{''}} - 4 y^{^{'}} + 4 y = 12 x + 7

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 3 x + \frac{19}{4}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 12 x + 7

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 3 x + \frac{19}{4}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 3 x + \frac{19}{4}

y^{^{''}} - 4 y^{^{'}} + 4 y = 12 x + 4

y^{^{''}} + y = sin (2 x) + cos (3 x)

\frac{1}{2} y^{^{''}} + \frac{1}{2} y^{^{'}} + 6 y = 10 cos (3 t)

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

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

What is the solution for y^{''}-4y^'+4y=12x+7 ?
The solution for y^{''}-4y^'+4y=12x+7 is y=c_{1}e^{2x}+c_{2}xe^{2x}+3x+19/4