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Popular >

y^'=-0.02(y-20)

Solution

y^{^{'}} = - 0.02 (y - 20)

Solution

y = e^{- 0.02 t + c_{1}} + 20

Solution steps

y^{'} (t) = - 0.02 (y - 20)

Solve separable ODE:

y = e^{- 0.02 t + c_{1}} + 20

y = e^{- 0.02 t + c_{1}} + 20

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'=-0.02(y-20) ?
The solution for y^'=-0.02(y-20) is y=e^{-0.02t+c_{1}}+20