What is the solution for y^{''}+4y=x^2+1 ?

The solution for y^{''}+4y=x^2+1 is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2)/4+1/8

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y^{''}+4y=x^2+1

y^{^{''}} + 4 y = x^{2} + 1

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} + \frac{1}{8}

Solution steps

y^{''} (x) + 4 y = x^{2} + 1

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} + \frac{1}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} + \frac{1}{8}

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} + 6 y = e^{x}

y^{^{''}} + y = 12 cos^{2} (x)

y^{^{''}} + y^{^{'}} - 2 y = t^{2}

x^{^{''}} + 3 x^{^{'}} + 33 x = 10 cos (5 t)

y^{^{''}} + 4 y^{^{'}} + 3 y = 5 e^{- 3 t}

What is the solution for y^{''}+4y=x^2+1 ?
The solution for y^{''}+4y=x^2+1 is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2)/4+1/8