What is the solution for y^{''}-2y^'+3y=e^x ?

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

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

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

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 3 y = e^{x}

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} = 8 e^{2 t} - 2 e^{- t}

y^{^{''}} - 2 y^{^{'}} + y = t

y^{^{''}} - 4 y^{^{'}} = 8 e^{6 x} (x - 1)

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

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 8 e^{x}

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