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

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

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y^{''}+y^'=2cos(t)

y^{^{''}} + y^{^{'}} = 2 cos (t)

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

Solution steps

y^{''} (t) + y^{'} (t) = 2 cos (t)

Solve linear ODE:

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

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

y^{^{''}} + 1000 y^{^{'}} + 2470 = 0

x^{^{''}} + 2 x^{^{'}} + 1 x = e^{- 2 t}, x (0) = 2, x^{^{'}} (0) = 1

y^{^{''}} + 4 y = cos (2 x) + 1

y^{^{''}} + 9 y = sin (3 x), y (0) = 3, y^{^{'}} (0) = - \frac{1}{6}

\frac{d ^{2} y}{d x ^{2}} + y = 4 x cos (x)

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