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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 = 0

y^{^{'''}} - 6 y^{^{''}} = 4 - sin (x)

y^{^{''}} + 3 y = 5, y (0) = 1, y^{^{'}} (0) = 1

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

y^{^{''}} + 2 y^{^{'}} + 2 y = a cos (x)

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