What is the solution for y^{''}+4y=4+2x ?

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

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x}{2} + 1

Solution steps

y^{''} (x) + 4 y = 4 + 2 x

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x}{2} + 1

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x}{2} + 1

2 x^{^{''}} + 32 x = 68 e^{- 2 t} cos (4 t)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} = sin (x) - cos (x)

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

y^{^{''}} + 16 y = 56 cos (4 t)

x^{^{''}} + 5 x^{^{'}} + 17 x = 12 cos (3 t)

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