What is the solution for y^{''}+4y^'+3y=65sin(2t) ?

The solution for y^{''}+4y^'+3y=65sin(2t) is y=c_{1}e^{-t}+c_{2}e^{-3t}-sin(2t)-8cos(2t)

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y^{''}+4y^'+3y=65sin(2t)

y^{^{''}} + 4 y^{^{'}} + 3 y = 65 sin (2 t)

y = c_{1} e^{- t} + c_{2} e^{- 3 t} - sin (2 t) - 8 cos (2 t)

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 65 sin (2 t)

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} - sin (2 t) - 8 cos (2 t)

y = c_{1} e^{- t} + c_{2} e^{- 3 t} - sin (2 t) - 8 cos (2 t)

f (x) = 3 - x

x^{2} + 1, x, [0, 1]

\frac{\partial}{\partial x} (3 x y^{2} - 2 y + 5 x^{2} y^{2})

f (x) = x + 3, at x = - 2

s^{2}

What is the solution for y^{''}+4y^'+3y=65sin(2t) ?
The solution for y^{''}+4y^'+3y=65sin(2t) is y=c_{1}e^{-t}+c_{2}e^{-3t}-sin(2t)-8cos(2t)