What is the solution for y^{''}-4y=sin(2t) ?

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

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y^{''}-4y=sin(2t)

y^{^{''}} - 4 y = sin (2 t)

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{8} sin (2 t)

Solution steps

y^{''} (t) - 4 y = sin (2 t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{8} sin (2 t)

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{8} sin (2 t)

\int (x)^{\frac{1}{2}} d x

y = - 5 - 8 x^{2}, (5, - 205)

\frac{2}{1 - x}, 0

\frac{1}{1 + x ^{2}}, 0

\int_{0}^{1} x^{2} (2 x^{3} - 1)^{4} d x

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