What is the solution for y^{''}+10y^'+25y=x,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+10y^'+25y=x,y(0)=0,y^'(0)=0 is y= 2/125 e^{-5x}+(e^{-5x}x)/(25)+x/(25)-2/125

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y^{''}+10y^'+25y=x,y(0)=0,y^'(0)=0

y^{^{''}} + 10 y^{^{'}} + 25 y = x, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{2}{125} e^{- 5 x} + \frac{e ^{- 5 x} x}{25} + \frac{x}{25} - \frac{2}{125}

Solution steps

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

Solve linear ODE:

y = \frac{2}{125} e^{- 5 x} + \frac{e ^{- 5 x} x}{25} + \frac{x}{25} - \frac{2}{125}

y = \frac{2}{125} e^{- 5 x} + \frac{e ^{- 5 x} x}{25} + \frac{x}{25} - \frac{2}{125}

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

3 y^{^{''}} + 27 y = 2 cos (3 x)

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 cos (2 t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + y = 3 cos (t)

What is the solution for y^{''}+10y^'+25y=x,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+10y^'+25y=x,y(0)=0,y^'(0)=0 is y= 2/125 e^{-5x}+(e^{-5x}x)/(25)+x/(25)-2/125