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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

\frac{d ^{2} x}{d t ^{2}} + x k + a = 0

y^{^{''}} - 2 y^{^{'}} + y = x^{\frac{1}{2}} e^{x}

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

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