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

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

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

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

y = \frac{1}{5} e^{- x} + \frac{2}{5} e^{x} sin (x) - \frac{1}{5} e^{x} cos (x)

Solution steps

y^{''} (x) - y = e^{x} cos (x)

Solve linear ODE:

y = \frac{1}{5} e^{- x} + \frac{2}{5} e^{x} sin (x) - \frac{1}{5} e^{x} cos (x)

y = \frac{1}{5} e^{- x} + \frac{2}{5} e^{x} sin (x) - \frac{1}{5} e^{x} cos (x)

y^{^{''}} - 6 y = e^{x}

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 6} e^{5 t}

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

q^{^{''}} + 100 q^{^{'}} + 2500 q = 40, q (0) = 0

(D^{2} + 5) y = cos (5) x

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