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Popular >

t^2y^'-y=0

Solution

t^{2} y^{^{'}} - y = 0

Solution

y = e^{- \frac{1}{t} + c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{- \frac{1}{t} + c_{1}}

y = e^{- \frac{1}{t} + c_{1}}

Graph

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Frequently Asked Questions (FAQ)

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