Help solving nonlinear differential equation?

Invert both sides:

dt/dy = t + y

==> dt/dy - t = y.

This is linear in t, with integrating factor e^(-∫ 1 dy) = e^(-y):

e^(-y) dt/dy - e^(-y) t = ye^(-y)

==> (d/dt) [-e^(-y) t] = ye^(-y)

Integrate both sides:

-e^(-y) t = -ye^(-y) - e^(-y) + C

==> t = -y - 1 + Ce^y

Find C via (t, y) = (-1, 0):

-1 = -0 - 1 + Ce^0

==> C = 0.

Hence, the desired solution is t = -y - 1.

==> y = -t - 1.

I hope this helps!

Use an integrating factor to find the equation for the general solution:

dy / dt = 1 / (t + y)

dt / dy = t + y

dt / dy - t = y

dt / dy + P(y)t = f(y)

P(y) = -1

f(y) = y

I(y) = e ^ ∫ P(y) dy

I(y) = e ^ ∫ -1 dy

I(y) = e ^ (-y)

I(y)t = ∫ I(y)f(y) dy

te ^ (-y) = ∫ ye ^ (-y) dy

Integrate this function on the right side by parts:

∫ ye ^ (-y) dy

Let f'(y) = e ^ (-y)

f(y) = -e ^ (-y)

Let g(y) = y

g'(y) = 1

∫ f'(y)g(y) dy = f(y)g(y) - ∫ f(y)g'(y) dy

∫ ye ^ (-y) dy = -ye ^ (-y) + ∫ e ^ (-y) dy

∫ ye ^ (-y) dy = -ye ^ (-y) - e ^ (-y) + C

∫ ye ^ (-y) dy = -(y + 1)e ^ (-y) + C

∫ ye ^ (-y) dy = -(y + 1) / e ^ y + C

Now plug this integral into the equation to get the general solution:

te ^ (-y) = ∫ ye ^ (-y) dy

te ^ (-y) = -(y + 1) / e ^ y + C

t = -(y + 1) + Ce ^ y

t = Ce ^ y - y - 1

Finally solve for the constant to find the particular solution:

When t = -1, y = 0

-1 = C - 1

C = 0

t = -y - 1

y = -t - 1