What is the solution for 5y^{''}+y^'=-8x,y(0)=0,y^'(0)=-10 ?

The solution for 5y^{''}+y^'=-8x,y(0)=0,y^'(0)=-10 is y=-250+250e^{-x/5}-4x^2+40x

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5y^{''}+y^'=-8x,y(0)=0,y^'(0)=-10

5 y^{^{''}} + y^{^{'}} = - 8 x, y (0) = 0, y^{^{'}} (0) = - 10

y = - 250 + 250 e^{- \frac{x}{5}} - 4 x^{2} + 40 x

Solution steps

5 y^{''} (x) + y^{'} (x) = - 8 x

Solve linear ODE:

y = - 250 + 250 e^{- \frac{x}{5}} - 4 x^{2} + 40 x

y = - 250 + 250 e^{- \frac{x}{5}} - 4 x^{2} + 40 x

y^{^{''}} - 36 y = - 42 e^{- 5 t}, y (0) = 7, y^{^{'}} (0) = 0

y^{^{''}} - 7 y^{^{'}} + 6 y = 2, y^{^{'}} (0) = 7

y^{^{'''}} - y^{^{'}} = 1 - x^{2}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{(3 t)}

y^{^{''}} + 4 y = 3 x

What is the solution for 5y^{''}+y^'=-8x,y(0)=0,y^'(0)=-10 ?
The solution for 5y^{''}+y^'=-8x,y(0)=0,y^'(0)=-10 is y=-250+250e^{-x/5}-4x^2+40x