What is the solution for 5y^{''}+y^'=-4x,y(0)=0,y^'(0)=-25 ?

The solution for 5y^{''}+y^'=-4x,y(0)=0,y^'(0)=-25 is y=-225+225e^{-x/5}-2x^2+20x

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5y^{''}+y^'=-4x,y(0)=0,y^'(0)=-25

5 y^{^{''}} + y^{^{'}} = - 4 x, y (0) = 0, y^{^{'}} (0) = - 25

y = - 225 + 225 e^{- \frac{x}{5}} - 2 x^{2} + 20 x

Solution steps

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

Solve linear ODE:

y = - 225 + 225 e^{- \frac{x}{5}} - 2 x^{2} + 20 x

y = - 225 + 225 e^{- \frac{x}{5}} - 2 x^{2} + 20 x

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} = 5 t

3 y^{^{''}} + 2 y^{^{'}} - y = 23

y^{^{''}} - y^{^{'}} - 6 y = 45 cos (3 x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 8 e^{- 2 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 e^{2 t}

What is the solution for 5y^{''}+y^'=-4x,y(0)=0,y^'(0)=-25 ?
The solution for 5y^{''}+y^'=-4x,y(0)=0,y^'(0)=-25 is y=-225+225e^{-x/5}-2x^2+20x