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

The solution for y^{''}-2y^'+5y=e^x,y(0)=0,y^'(0)=0 is y=-1/4 e^xcos(2x)+1/4 e^x

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

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

y = - \frac{1}{4} e^{x} cos (2 x) + \frac{1}{4} e^{x}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{4} e^{x} cos (2 x) + \frac{1}{4} e^{x}

y = - \frac{1}{4} e^{x} cos (2 x) + \frac{1}{4} e^{x}

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{(- 3)} e^{3} x

x^{^{''}} + 2 x + 3 x^{^{'}} = 5 cos (t) + sin (2 t)

(D^{2} + 2 D + 1) y = e^{- x}

y^{^{''}} - y^{^{'}} = 4 x e^{x}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x}, y (0) = - 1, y^{^{'}} (0) = 1

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