What is the solution for y^{''}+2y^'=x+1 ?

The solution for y^{''}+2y^'=x+1 is y=c_{1}+c_{2}e^{-2x}+(x^2)/4+x/4

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

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

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{2}}{4} + \frac{x}{4}

Solution steps

y^{''} (x) + 2 y^{'} (x) = x + 1

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{2}}{4} + \frac{x}{4}

y = c_{1} + c_{2} e^{- 2 x} + \frac{x ^{2}}{4} + \frac{x}{4}

x^{^{''}} + x = 8 cos (6 t)

y^{^{''}} + 2 y^{^{'}} - 3 y = x^{2} - 1

\frac{d ^{2} x}{d t ^{2}} + 16 x = 12 cos (9 t)

y^{^{''}} - 3 y^{^{'}} - 18 y = x e^{3 x}

y^{^{''}} + y = e^{- t} sin (e^{- t}) + cos (e^{- t})

What is the solution for y^{''}+2y^'=x+1 ?
The solution for y^{''}+2y^'=x+1 is y=c_{1}+c_{2}e^{-2x}+(x^2)/4+x/4