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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 2 x = 1

x^{^{''}} + 2 x = 6

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 2 y = 2 x^{2} - 3 x + 6

y^{^{''}} + 10 y^{^{''}} + 33 y^{^{'}} + 36 y = 4 t e^{- 3 t}

x^{^{''}} + 4 x = 12 cos (2 t)

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