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Popular >

tv^'(1-t^2)+2v=0

Solution

t v^{^{'}} (1 - t^{2}) + 2 v = 0

Solution

v (t) = \frac{c _{1}}{t ^{2}}

Solution steps

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

Solve separable ODE:

v (t) = \frac{c _{1}}{t ^{2}}

v (t) = \frac{c _{1}}{t ^{2}}

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\int (x^{5} + 1) d x

Frequently Asked Questions (FAQ)

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