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

The solution for y^{''}-2y^'+4y=e^x is y=e^x(c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x))+1/3 e^x

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

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

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{1}{3} e^{x}

Solution steps

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

Solve linear ODE:

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{1}{3} e^{x}

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{1}{3} e^{x}

What is the solution for y^{''}-2y^'+4y=e^x ?
The solution for y^{''}-2y^'+4y=e^x is y=e^x(c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x))+1/3 e^x