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Popular >

y^'=(t+y)^2

Solution

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

Solution

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

Solution steps

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

Solve using substitution:

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

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

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Frequently Asked Questions (FAQ)

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