What is the solution for y^'-y=4sin(3t) ?

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

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y^'-y=4sin(3t)

y^{^{'}} - y = 4 sin (3 t)

y = \frac{2 e ^{t} ( - 3 e ^{- t} cos ( 3 t ) - e ^{- t} sin ( 3 t ) )}{5} + c_{1} e^{t}

Solution steps

y^{'} (t) - y = 4 sin (3 t)

Solve linear ODE:

y = \frac{2 e ^{t} ( - 3 e ^{- t} cos ( 3 t ) - e ^{- t} sin ( 3 t ) )}{5} + c_{1} e^{t}

y = \frac{2 e ^{t} ( - 3 e ^{- t} cos ( 3 t ) - e ^{- t} sin ( 3 t ) )}{5} + c_{1} e^{t}

\frac{1}{s - 1}

n = 4 \sum \infty \frac{1}{( n ) ( n + 1 )}

\int_{0}^{x} \frac{1}{( 1 - x )} d x

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

n = 1 \sum \infty (cos (32))^{n}

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