What is the solution for y^{''}+4y=8sin(2x)+5e^x ?

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

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y^{''}+4y=8sin(2x)+5e^x

y^{^{''}} + 4 y = 8 sin (2 x) + 5 e^{x}

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

Solution steps

y^{''} (x) + 4 y = 8 sin (2 x) + 5 e^{x}

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12 x, y (0) = 1, y^{^{'}} (0) = - 2

y^{^{''}} + 25 y = 82 e^{4 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = - 3 e^{2 x} + \frac{e ^{2 x}}{x}

y^{^{''}} + 8 y^{^{'}} + 12 y = 1 - H (t - 3)

y^{^{''}} - 6 y^{^{'}} + 9 y = 2, y^{^{'}} (0) = 6

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