What is the solution for y^{''}-3y^'-4y=2sin(t) ?

The solution for y^{''}-3y^'-4y=2sin(t) is y=c_{1}e^{4t}+c_{2}e^{-t}-5/17 sin(t)+3/17 cos(t)

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y^{''}-3y^'-4y=2sin(t)

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 sin (t)

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{5}{17} sin (t) + \frac{3}{17} cos (t)

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 2 sin (t)

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{5}{17} sin (t) + \frac{3}{17} cos (t)

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{5}{17} sin (t) + \frac{3}{17} cos (t)

\frac{d}{d x} (5 x cos (x))

x \to 0 + lim (3 - x)

\frac{d}{d x} (3 sin (x) + 3 cos (x))

\int ln (x + 16) d x

\frac{d}{d x} (ln (x) \cdot ln (x))

What is the solution for y^{''}-3y^'-4y=2sin(t) ?
The solution for y^{''}-3y^'-4y=2sin(t) is y=c_{1}e^{4t}+c_{2}e^{-t}-5/17 sin(t)+3/17 cos(t)