What is the solution for sin(t)=y^'+y ?

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

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sin(t)=y^'+y

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

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

Solution steps

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

Solve linear ODE:

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

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

n = 1 \sum \infty \frac{- 1}{n + 1}

\int (1 - 6 x) e^{3 x - 9 x^{2}} d x

n \to \infty lim (7^{n})

\int_{0}^{11} 2 π y (121 - y^{2}) d y

\frac{\partial}{\partial y} (x^{2} + y - e^{x + y})

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