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Popular Calculus >

y^'=-1/40 y

Solution

y^{^{'}} = - \frac{1}{40} y

Solution

y = e^{\frac{- t - c _{1}}{40}}

Solution steps

y^{'} (t) = - \frac{1}{40} y

Solve separable ODE:

y = e^{\frac{- t - c _{1}}{40}}

y = e^{\frac{- t - c _{1}}{40}}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'=-1/40 y ?
The solution for y^'=-1/40 y is y=e^{(-t-c_{1})/(40)}