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

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

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

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

y = c_{1} cos (x) + c_{2} sin (x) + 2 x + 4

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + 2 x + 4

y = c_{1} cos (x) + c_{2} sin (x) + 2 x + 4

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = sin (x)

y^{^{''}} + 16 y = - 16 sec (4 t)

\frac{d ^{2} y}{d x ^{2}} - 10 \cdot \frac{d y}{d x} + 25 \cdot y = 30 \cdot x + 3

y^{^{''}} - 5 y^{^{'}} = 28 x e^{- 2 x} + 10

y^{^{''}} - 9 y^{^{'}} + 20 y = 6 x + e^{4 x}

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