What is the solution for 2y^'+y=0,y(0)=e^{(-x)/2} ?

The solution for 2y^'+y=0,y(0)=e^{(-x)/2} is y=e^{(-x+c_{1})/2}

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2y^'+y=0,y(0)=e^{(-x)/2}

2 y^{^{'}} + y = 0, y (0) = e^{\frac{- x}{2}}

y = e^{\frac{- x + c _{1}}{2}}

Solution steps

2 y^{'} (x) + y = 0

Solve separable ODE:

y = e^{\frac{- x + c _{1}}{2}}

y = e^{\frac{- x + c _{1}}{2}}

y^{^{'}} = \frac{x ^{2} - 6}{y ^{2}}, y (0) = 2

\frac{d y}{d x} = \frac{x ^{2}}{x ^{2} ( 3 y ^{2} + 1 )}, y (1) = 4

\frac{d y}{d x} = x^{5} y^{2}

5 x y y^{^{'}} = 1

y^{^{'}} = \frac{2 y}{t - 1}

What is the solution for 2y^'+y=0,y(0)=e^{(-x)/2} ?
The solution for 2y^'+y=0,y(0)=e^{(-x)/2} is y=e^{(-x+c_{1})/2}