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

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

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{2}{5} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{2}{5} e^{x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{2}{5} e^{x}

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