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Popular >

y^'+8y=0

Solution

y^{^{'}} + 8 y = 0

Solution

y = e^{- 8 t + c_{1}}

Solution steps

y^{'} (t) + 8 y = 0

Solve separable ODE:

y = e^{- 8 t + c_{1}}

y = e^{- 8 t + c_{1}}

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\int_{0}^{3} π (2 7 y)^{2} d y

derivative of 1/((3^{(-x^2-y^2)+1)})

\frac{d}{d x} (\frac{1}{( 3 ^{(- x^{2} - y^{2})} + 1 )})

Frequently Asked Questions (FAQ)

What is the solution for y^'+8y=0 ?
The solution for y^'+8y=0 is y=e^{-8t+c_{1}}