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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to - 6 lim (\frac{x ^{2} - 4}{6 - x})

\int 2 t - 1 d t

4 x + 9

\int (4 x - 3)^{5} d x

\int cos (n wt) d t

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