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= 1/5 e^{-t}+2/5 e^tsin(t)-1/5 e^tcos(t)

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 1, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + y = t^{- 4} e^{t}

y^{^{''}} + y^{^{'}} = 10 x^{4} + 2

y^{^{''}} - 2 y^{^{'}} - 3 y = x - e^{x}

x^{^{''}} - 4 x = 3 t

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= 1/5 e^{-t}+2/5 e^tsin(t)-1/5 e^tcos(t)