What is the solution for y^'-y=e^{-t}sin(t) ?

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

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y^'-y=e^{-t}sin(t)

y^{^{'}} - y = e^{- t} sin (t)

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

Solution steps

y^{'} (t) - y = e^{- t} sin (t)

Solve linear ODE:

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

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

n = 1 \sum \infty ln (n^{2} + 2 n)

n = 1 \sum \infty \frac{1}{7 ^{n}}

\int (x - 2)^{2} sin (x) d x

\int \frac{1}{x ^{2} + 3 x - 4} d x

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

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