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Popular >

(t^2-1)y^'+2ty=0

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

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Frequently Asked Questions (FAQ)

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