What is the solution for y^{''}+4y=x^2+5e^x ?

The solution for y^{''}+4y=x^2+5e^x is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2)/4-1/8+e^x

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y^{''}+4y=x^2+5e^x

y^{^{''}} + 4 y = x^{2} + 5 e^{x}

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

Solution steps

y^{''} (x) + 4 y = x^{2} + 5 e^{x}

Solve linear ODE:

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

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

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

y^{^{''}} + 6 y^{^{'}} + 3 x = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 5 y = 5 sin (x), y (0) = 1, y^{^{'}} (0) = 2

y^{^{'''}} + y^{^{'}} = sec (t), y^{^{'}} (0) = 3, y^{^{''}} (0) = - 2

\frac{d ^{2} y}{d x ^{2}} - 2 \cdot \frac{d y}{d x} + y = (x - 1) \cdot e^{x}

What is the solution for y^{''}+4y=x^2+5e^x ?
The solution for y^{''}+4y=x^2+5e^x is y=c_{1}cos(2x)+c_{2}sin(2x)+(x^2)/4-1/8+e^x