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

The solution for y+2y^'+5y=4*e^{-t}*cos(2t) is y=(sin(2t)+cos(2t))/(2e^t)+(c_{1})/(e^{3t)}

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

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

y = \frac{sin ( 2 t ) + cos ( 2 t )}{2 e ^{t}} + \frac{c _{1}}{e ^{3 t}}

Solution steps

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

Solve linear ODE:

y = \frac{sin ( 2 t ) + cos ( 2 t )}{2 e ^{t}} + \frac{c _{1}}{e ^{3 t}}

y = \frac{sin ( 2 t ) + cos ( 2 t )}{2 e ^{t}} + \frac{c _{1}}{e ^{3 t}}

\frac{x - 2}{x - 2}

u \to 2 + lim (u^{2})

\frac{d ^{2} y}{d x ^{2}}

\frac{\partial}{\partial x} (2 x e^{a y} a + 3 e^{x b})

What is the solution for y+2y^'+5y=4*e^{-t}*cos(2t) ?
The solution for y+2y^'+5y=4*e^{-t}*cos(2t) is y=(sin(2t)+cos(2t))/(2e^t)+(c_{1})/(e^{3t)}