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Popular >

2y^'+y=0

Solution

2 y^{^{'}} + y = 0

Solution

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

Solution steps

2 y^{'} (t) + y = 0

Solve separable ODE:

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

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

Graph

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Frequently Asked Questions (FAQ)

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