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Popular >

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

Solution

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

Solution

y = - \frac{1}{t + c _{1}} - t

Solution steps

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

Solve using substitution:

y = - \frac{1}{t + c _{1}} - t

y = - \frac{1}{t + c _{1}} - t

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Frequently Asked Questions (FAQ)

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