What is the solution for y^'-y=2cos(5t) ?

The solution for y^'-y=2cos(5t) is y=(e^t(5e^{-t}sin(5t)-e^{-t}cos(5t)))/(13)+c_{1}e^t

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y^'-y=2cos(5t)

y^{^{'}} - y = 2 cos (5 t)

y = \frac{e ^{t} ( 5 e ^{- t} sin ( 5 t ) - e ^{- t} cos ( 5 t ) )}{13} + c_{1} e^{t}

Solution steps

y^{'} (t) - y = 2 cos (5 t)

Solve linear ODE:

y = \frac{e ^{t} ( 5 e ^{- t} sin ( 5 t ) - e ^{- t} cos ( 5 t ) )}{13} + c_{1} e^{t}

y = \frac{e ^{t} ( 5 e ^{- t} sin ( 5 t ) - e ^{- t} cos ( 5 t ) )}{13} + c_{1} e^{t}

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

x \to 0 lim (\frac{1}{2 ^{x} - 1} - \frac{1}{x})

f (x) = \frac{1}{x ^{5}}, at x = 1

\int \frac{t ^{6}}{1 - t ^{14}} d t

x \to a + lim (g (x))

What is the solution for y^'-y=2cos(5t) ?
The solution for y^'-y=2cos(5t) is y=(e^t(5e^{-t}sin(5t)-e^{-t}cos(5t)))/(13)+c_{1}e^t