What is the solution for 1x^{''}+2x=8sin(2t) ?

The solution for 1x^{''}+2x=8sin(2t) is x=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)-4sin(2t)

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1x^{''}+2x=8sin(2t)

1 x^{^{''}} + 2 x = 8 sin (2 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) - 4 sin (2 t)

Solution steps

1 x^{''} (t) + 2 x = 8 sin (2 t)

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) - 4 sin (2 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) - 4 sin (2 t)

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

y^{^{''}} - 8 y^{^{'}} + 16 = 10 e^{4 x} sec (x)

y^{^{''}} + 6 y^{^{'}} + 5 y = 4 t e^{4 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{3 t} ln (t)

y^{^{''}} - y = e^{2 x}, y (0) = 1

What is the solution for 1x^{''}+2x=8sin(2t) ?
The solution for 1x^{''}+2x=8sin(2t) is x=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)-4sin(2t)