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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 9 y^{^{''}} - 108 y = e^{- 6 x}

y^{^{''''}} + 2 y^{^{''}} + y = 3 t + 4

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

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 5 y = e^{x} cos (2 x)

4 y^{^{''}} - 4 y^{^{'}} + y = 32 e^{\frac{t}{2}}

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