What is the solution for y^{''}+4y^'+5=50x+13e^{3x} ?

The solution for y^{''}+4y^'+5=50x+13e^{3x} is y=c_{1}+c_{2}e^{-4x}+(1050x^2-735x+104e^{3x})/(168)

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y^{''}+4y^'+5=50x+13e^{3x}

y^{^{''}} + 4 y^{^{'}} + 5 = 50 x + 13 e^{3 x}

y = c_{1} + c_{2} e^{- 4 x} + \frac{1050 x ^{2} - 735 x + 104 e ^{3 x}}{168}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 = 50 x + 13 e^{3 x}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{1050 x ^{2} - 735 x + 104 e ^{3 x}}{168}

y = c_{1} + c_{2} e^{- 4 x} + \frac{1050 x ^{2} - 735 x + 104 e ^{3 x}}{168}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} = 12 x

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

y^{^{''}} - y^{^{'}} - 2 y = - 4 x^{2} - 6 x + 1

y^{^{''}} - 4 y^{^{'}} + 5 y = 10 x - 3

y^{^{''}} - 6 y^{^{'}} + 8 y = x e^{x}

What is the solution for y^{''}+4y^'+5=50x+13e^{3x} ?
The solution for y^{''}+4y^'+5=50x+13e^{3x} is y=c_{1}+c_{2}e^{-4x}+(1050x^2-735x+104e^{3x})/(168)