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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 lim (- 5^{2})

\frac{d}{d x} (x lo g_{10} (x))

\int y sin (x y) d x

\frac{d}{d x} (- 4 x^{10})

\frac{d y}{d t} = - 3 y + 12

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