What is the solution for y^'+y=e^{-t}sin(t),y(0)=1 ?

The solution for y^'+y=e^{-t}sin(t),y(0)=1 is y=-(cos(t))/(e^t)+2/(e^t)

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y^'+y=e^{-t}sin(t),y(0)=1

y^{^{'}} + y = e^{- t} sin (t), y (0) = 1

y = - \frac{cos ( t )}{e ^{t}} + \frac{2}{e ^{t}}

Solution steps

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

Solve linear ODE:

y = - \frac{cos ( t )}{e ^{t}} + \frac{2}{e ^{t}}

y = - \frac{cos ( t )}{e ^{t}} + \frac{2}{e ^{t}}

y^{^{'}} = y (1 - y)

x \to - 4 lim (8 x^{2} + 1)

- 2 sin (4 x)

y = ln (x^{8})

\int - csc^{2} (\frac{8 x}{5}) d x

What is the solution for y^'+y=e^{-t}sin(t),y(0)=1 ?
The solution for y^'+y=e^{-t}sin(t),y(0)=1 is y=-(cos(t))/(e^t)+2/(e^t)