What is the solution for y^{''}+2y^'+y=5sin(t) ?

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

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y^{''}+2y^'+y=5sin(t)

y^{^{''}} + 2 y^{^{'}} + y = 5 sin (t)

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 5 sin (t)

Solve linear ODE:

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

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

\int_{1}^{27} \frac{4 x ^{\frac{2}{3}} + 1}{4 x ^{\frac{1}{3}}} d x

e^{x}, x^{2} - 1, [- 1, 1]

\int 2 x sin (3 x) d x

y = x^{2} - 13, y = 5 - 3 x

x^{2} + x, at x = - 1

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