What is the solution for (dy)/(dt)=-3y+4cos(2t) ?

The solution for (dy)/(dt)=-3y+4cos(2t) is y=(4(2sin(2t)+3cos(2t)))/(13)+(c_{1})/(e^{3t)}

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(dy)/(dt)=-3y+4cos(2t)

\frac{d y}{d t} = - 3 y + 4 cos (2 t)

y = \frac{4 ( 2 sin ( 2 t ) + 3 cos ( 2 t ) )}{13} + \frac{c _{1}}{e ^{3 t}}

Solution steps

\frac{d y}{d t} = - 3 y + 4 cos (2 t)

Solve linear ODE:

y = \frac{4 ( 2 sin ( 2 t ) + 3 cos ( 2 t ) )}{13} + \frac{c _{1}}{e ^{3 t}}

y = \frac{4 ( 2 sin ( 2 t ) + 3 cos ( 2 t ) )}{13} + \frac{c _{1}}{e ^{3 t}}

(7 x y^{2} - 2 y) + (7 x^{2} y - 2 x) y^{^{'}} = 0

\int 20 x d x

\int \frac{x + 1}{( 2 - x ) ^{2}} d x

\int_{0}^{4} (e^{- 2 x} + 1) d x

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

What is the solution for (dy)/(dt)=-3y+4cos(2t) ?
The solution for (dy)/(dt)=-3y+4cos(2t) is y=(4(2sin(2t)+3cos(2t)))/(13)+(c_{1})/(e^{3t)}