What is the solution for y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0 ?

The solution for y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0 is y=-3cos(t)+3sin(t)+3e^{-t}

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y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0

y^{^{''}} + y = 6 e^{- t}, y^{^{'}} (0) = 0, y (0) = 0

y = - 3 cos (t) + 3 sin (t) + 3 e^{- t}

Solution steps

y^{''} (t) + y = 6 e^{- t}

Solve linear ODE:

y = - 3 cos (t) + 3 sin (t) + 3 e^{- t}

y = - 3 cos (t) + 3 sin (t) + 3 e^{- t}

What is the solution for y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0 ?
The solution for y^{''}+y=6e^{-t},y^'(0)=0,y(0)=0 is y=-3cos(t)+3sin(t)+3e^{-t}