What is the solution for y^{''}-y^'=e^tcos(t),y(0)=0,y^'(0)=0 ?

The solution for y^{''}-y^'=e^tcos(t),y(0)=0,y^'(0)=0 is y= 1/2 e^tsin(t)-1/2 e^tcos(t)+1/2

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y^{''}-y^'=e^tcos(t),y(0)=0,y^'(0)=0

y^{^{''}} - y^{^{'}} = e^{t} cos (t), y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) - y^{'} (t) = e^{t} cos (t)

Solve linear ODE:

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

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

What is the solution for y^{''}-y^'=e^tcos(t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}-y^'=e^tcos(t),y(0)=0,y^'(0)=0 is y= 1/2 e^tsin(t)-1/2 e^tcos(t)+1/2