What is the solution for y^{''}+10y^'+25y=30x+3 ?

The solution for y^{''}+10y^'+25y=30x+3 is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+(6x)/5-9/25

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y^{''}+10y^'+25y=30x+3

y^{^{''}} + 10 y^{^{'}} + 25 y = 30 x + 3

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{6 x}{5} - \frac{9}{25}

Solution steps

y^{''} (x) + 10 y^{'} (x) + 25 y = 30 x + 3

Solve linear ODE:

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{6 x}{5} - \frac{9}{25}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{6 x}{5} - \frac{9}{25}

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

y^{^{''}} + 9 y = \frac{3}{2} sin (2 t)

y^{^{''}} - 2 y^{^{'}} - 8 y = 10 e^{3 x}

y^{^{''}} + 2 y^{^{'}} - 8 y = 5 e^{- 2 x} - e^{- x}

y^{^{''}} - y^{^{'}} - 2 y = 5 x + 3

What is the solution for y^{''}+10y^'+25y=30x+3 ?
The solution for y^{''}+10y^'+25y=30x+3 is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+(6x)/5-9/25