What is the solution for y^{''}-y^'-30y=2x+4 ?

The solution for y^{''}-y^'-30y=2x+4 is y=c_{1}e^{6x}+c_{2}e^{-5x}-x/(15)-59/450

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y^{''}-y^'-30y=2x+4

y^{^{''}} - y^{^{'}} - 30 y = 2 x + 4

y = c_{1} e^{6 x} + c_{2} e^{- 5 x} - \frac{x}{15} - \frac{59}{450}

Solution steps

y^{''} (x) - y^{'} (x) - 30 y = 2 x + 4

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{- 5 x} - \frac{x}{15} - \frac{59}{450}

y = c_{1} e^{6 x} + c_{2} e^{- 5 x} - \frac{x}{15} - \frac{59}{450}

x^{^{''}} + 4 x = 1

y^{^{''}} + y = 5 x sin (5 x)

(D^{4} + 2 D^{3} + 2 D^{2}) y = 2 x e^{- x}

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 12 = 0

y^{^{''}} + 4 y^{^{'}} + 5 y = \frac{2 e ^{2 x}}{sin ( x )}

What is the solution for y^{''}-y^'-30y=2x+4 ?
The solution for y^{''}-y^'-30y=2x+4 is y=c_{1}e^{6x}+c_{2}e^{-5x}-x/(15)-59/450