What is the solution for y^{''}-y=4sin(3t) ?

The solution for y^{''}-y=4sin(3t) is y=c_{1}e^t+c_{2}e^{-t}-2/5 sin(3t)

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y^{''}-y=4sin(3t)

y^{^{''}} - y = 4 sin (3 t)

y = c_{1} e^{t} + c_{2} e^{- t} - \frac{2}{5} sin (3 t)

Solution steps

y^{''} (t) - y = 4 sin (3 t)

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} - \frac{2}{5} sin (3 t)

y = c_{1} e^{t} + c_{2} e^{- t} - \frac{2}{5} sin (3 t)

\int - 2 x^{4} - 4 x^{2} - 5 d x

\frac{d}{d x} (cot (\frac{π}{2}))

\frac{d ^{2} y}{d x ^{2}} (x^{2}) - 9 x \frac{d y}{d x} + (61 y) = 0

x \to - 1 lim (2 x^{2} + 3 x - 5)

(f (6 x))^{2} + 3

What is the solution for y^{''}-y=4sin(3t) ?
The solution for y^{''}-y=4sin(3t) is y=c_{1}e^t+c_{2}e^{-t}-2/5 sin(3t)