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

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

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x^'+2x=e^{-2t}cos(t)

x^{^{'}} + 2 x = e^{- 2 t} cos (t)

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

Solution steps

x^{'} (t) + 2 x = e^{- 2 t} cos (t)

Solve linear ODE:

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

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

\int y ln (y) d y

f (x) = x + 1

\frac{d}{d x} (3 x y^{2})

\frac{d}{d x} (\frac{sin ( y )}{sin ( x )})

y (θ) = \frac{π}{8} sin (θ)

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