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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 6 cos (4 x) + 9 sin (4 x)

y^{^{'''}} - 2 y^{^{''}} + 4 y^{^{'}} - 8 = 0

y^{^{''''}} - y = 12 t + cos (t)

y^{^{''}} + 8 y^{^{'}} = 8 x

y^{^{''}} + 2 y^{^{'}} + 2 y = x^{2} - 3 x e^{- 2 x} cos (5 x)

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