What is the solution for y^{''}+2y^'+5y=4e^{-t}cos(2t) ?

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

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y^{''}+2y^'+5y=4e^{-t}cos(2t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{- t} cos (2 t)

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 4 e^{- t} cos (2 t)

Solve linear ODE:

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

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

x \cdot d x - y^{2} \cdot d y = 0

y^{^{'}} = x^{2} + 1

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{15}{2}}} d t

\int (x^{6} - 2 x^{3})^{5} (x^{5} - x^{2}) d x

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