What is the solution for y^{''}+4y^'+4y=3x+7 ?

The solution for y^{''}+4y^'+4y=3x+7 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3x)/4+1

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y^{''}+4y^'+4y=3x+7

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 x + 7

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 x}{4} + 1

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 3 x + 7

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 x}{4} + 1

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 x}{4} + 1

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

\frac{1}{2} y^{^{''}} + \frac{1}{2} y^{^{'}} + 6 y = 10 cos (3 x)

y^{^{''}} + \frac{750}{30} y = \frac{20}{30} cos (2) t

5 y^{^{''}} + 180 y = 150 sin (3 t)

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

What is the solution for y^{''}+4y^'+4y=3x+7 ?
The solution for y^{''}+4y^'+4y=3x+7 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3x)/4+1