What is the solution for y^'+y=u,y(0)=(-1)/2 ?

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

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y^'+y=u,y(0)=(-1)/2

y^{^{'}} + y = u, y (0) = \frac{- 1}{2}

y = - u e^{- t} - \frac{1}{2} e^{- t} + u

Solution steps

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

Solve separable ODE:

y = - u e^{- t} - \frac{1}{2} e^{- t} + u

y = - u e^{- t} - \frac{1}{2} e^{- t} + u

\frac{d y}{d x} = \frac{- x}{e ^{y}}

\frac{d x}{d t} = 14 - 0.7 x

\frac{d y}{d x} = \frac{3 sin ( x )}{y}

\frac{d y}{d x} + 16 x y = 0

2 x \frac{d y}{d x} + (1 - x) \frac{d y}{d x} - 2 y = 0

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