What is the solution for y^{''}-2y^'+5y=e^x(0=(2x)) ?

The solution for y^{''}-2y^'+5y=e^x(0=(2x)) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+(2x)/5+4/25

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y^{''}-2y^'+5y=e^x(0=(2x))

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x} (0 = (2 x))

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2 x}{5} + \frac{4}{25}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = e^{x} (0 = (2 x))

Solve linear ODE:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2 x}{5} + \frac{4}{25}

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2 x}{5} + \frac{4}{25}

What is the solution for y^{''}-2y^'+5y=e^x(0=(2x)) ?
The solution for y^{''}-2y^'+5y=e^x(0=(2x)) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+(2x)/5+4/25