What is the solution for y^{''}+4y=8e^{2t},y(0)=1,y^'(0)=-2 ?

The solution for y^{''}+4y=8e^{2t},y(0)=1,y^'(0)=-2 is y=e^{2t}-2sin(2t)

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y^{''}+4y=8e^{2t},y(0)=1,y^'(0)=-2

y^{^{''}} + 4 y = 8 e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 2

y = e^{2 t} - 2 sin (2 t)

Solution steps

y^{''} (t) + 4 y = 8 e^{2 t}

Solve linear ODE:

y = e^{2 t} - 2 sin (2 t)

y = e^{2 t} - 2 sin (2 t)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 2 x + 5 - e^{- 2 x}

D (x) = (x - 8)^{2}, S (x) = x^{2} + 2 x + 28

y^{^{''}} + 2 y^{^{'}} = x^{2} + 1

y^{^{''}} - y^{^{'}} - 20 y = 27 e^{- 4 c}

(D - 2)^{2} y = 8 e^{2 x}

What is the solution for y^{''}+4y=8e^{2t},y(0)=1,y^'(0)=-2 ?
The solution for y^{''}+4y=8e^{2t},y(0)=1,y^'(0)=-2 is y=e^{2t}-2sin(2t)