What is the solution for y^{''}+y^'+4=-20x ?

The solution for y^{''}+y^'+4=-20x is y=c_{1}+c_{2}e^{-x}-10x^2+16x

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y^{''}+y^'+4=-20x

y^{^{''}} + y^{^{'}} + 4 = - 20 x

y = c_{1} + c_{2} e^{- x} - 10 x^{2} + 16 x

Solution steps

y^{''} (x) + y^{'} (x) + 4 = - 20 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - 10 x^{2} + 16 x

y = c_{1} + c_{2} e^{- x} - 10 x^{2} + 16 x

y^{^{''}} - 6 y^{^{'}} + 9 y = 15 x + 3

y^{^{''}} - 6 y^{^{'}} + 9 y = 15 x + 2

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 15 = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{2} + 3 x

y^{^{''}} + 3 y^{^{'}} = 54 x

What is the solution for y^{''}+y^'+4=-20x ?
The solution for y^{''}+y^'+4=-20x is y=c_{1}+c_{2}e^{-x}-10x^2+16x