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Popular >

ty^'=y^2+1

Solution

t y^{^{'}} = y^{2} + 1

Solution

y = tan (ln (t) + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (ln (t) + c_{1})

y = tan (ln (t) + c_{1})

Graph

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Frequently Asked Questions (FAQ)

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