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Popular >

ty^'+2y=0

Solution

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

Solution

y = \frac{c _{1}}{t ^{2}}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{t ^{2}}

y = \frac{c _{1}}{t ^{2}}

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Frequently Asked Questions (FAQ)

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