What is the solution for (dy)/(dt)=(t+y)^2-1 ?

The solution for (dy)/(dt)=(t+y)^2-1 is y=-1/(t+c_{1)}-t

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(dy)/(dt)=(t+y)^2-1

\frac{d y}{d t} = (t + y)^{2} - 1

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

Solution steps

\frac{d y}{d t} = (t + y)^{2} - 1

Solve using substitution:

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

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

x \to 5 lim (9 - x^{2})

\int \frac{y}{y + 3} d y

x \to 0 lim (\frac{5 x ^{2}}{x ^{4}})

\int_{0}^{2} x^{3} 3 x + 1 d x

ln (x), at 4

What is the solution for (dy)/(dt)=(t+y)^2-1 ?
The solution for (dy)/(dt)=(t+y)^2-1 is y=-1/(t+c_{1)}-t