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= 1/2 e^xsin(x)-1/2 e^xcos(x)+1/2

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

Solution steps

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

Solve linear ODE:

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

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

(D^{2} + 5) y = 3 x e^{x}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 t} sin (t)

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{2} - x + 3

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

y^{^{''}} - 2 y^{^{'}} + y = - e^{x} sin (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= 1/2 e^xsin(x)-1/2 e^xcos(x)+1/2