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Popular >

y^'+0.5y=2

Solution

y^{^{'}} + 0.5 y = 2

Solution

y = - 2 e^{\frac{- t - c _{1}}{2}} + 4

Solution steps

y^{'} (t) + 0.5 y = 2

Solve separable ODE:

y = - 2 e^{\frac{- t - c _{1}}{2}} + 4

y = - 2 e^{\frac{- t - c _{1}}{2}} + 4

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'+0.5y=2 ?
The solution for y^'+0.5y=2 is y=-2e^{(-t-c_{1})/2}+4