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

The solution for y^'+y=e^{-t}sin(t) is y=-(cos(t))/(e^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 )}{e ^{t}} + \frac{c _{1}}{e ^{t}}

Solution steps

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

Solve linear ODE:

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

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

cos (t)

\frac{\partial}{\partial x} ((4 x^{4} y^{2} + 9)^{2})

n = 1 \sum \infty (\frac{7}{10})^{n}

\int \frac{1 + e ^{x}}{1 - e ^{x}} d x

h \to 0 lim (\frac{( 7 + h ) - 7}{h})

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