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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 9 y^{^{'}} + 14 y = 3 x^{2} - 5 sin (2 x) + 8 x e^{6 x}

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

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

y^{^{''}} - 6 y + 5 y^{^{'}} = - 9 e^{2 x}

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