What is the solution for y^{''}-2y^'=cos(t),y(0)=0,y^'(0)=1 ?

The solution for y^{''}-2y^'=cos(t),y(0)=0,y^'(0)=1 is y=(1/2+1/5)e^{2t}-1/2-2/5 sin(t)-1/5 cos(t)

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y^{''}-2y^'=cos(t),y(0)=0,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} = cos (t), y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) - 2 y^{'} (t) = cos (t)

Solve linear ODE:

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

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

What is the solution for y^{''}-2y^'=cos(t),y(0)=0,y^'(0)=1 ?
The solution for y^{''}-2y^'=cos(t),y(0)=0,y^'(0)=1 is y=(1/2+1/5)e^{2t}-1/2-2/5 sin(t)-1/5 cos(t)