I don't know this one:
$\displaystyle \frac{dy}{dt}=(y+t)^2$
Use substitution $\displaystyle z=y-t$ it says, but then $\displaystyle \frac{dy}{dt}=z^2$? Then what? Please
I expect you have to actually make the substitution $\displaystyle \displaystyle z = y + t$.
Then $\displaystyle \displaystyle y = z - t \implies \frac{dy}{dt} = \frac{dz}{dt} - 1$.
Substituting into the DE $\displaystyle \displaystyle \frac{dy}{dt} = (y + t)^2$ gives
$\displaystyle \displaystyle \begin{align*} \frac{dz}{dt} - 1 &= z^2 \\ \frac{dz}{dt} &= z^2 + 1 \\ \frac{1}{z^2 + 1}\,\frac{dz}{dt} &= 1 \\ \int{\frac{1}{z^2 + 1}\,\frac{dz}{dt}\,dt} &= \int{1\,dt} \\ \int{\frac{1}{z^2 + 1}\,dz} &= t + C_1 \\ \arctan{z} + C_2 &= t + C_1 \\ \arctan{z} &= t + C \textrm{ where }C = C_1 - C_2 \\ z &= \tan{(t + C)} \\ y + t &= \tan{(t + C)} \\ y &= \tan{(t + C)} - t \end{align*} $
Absolutely not. You can only use the method you have suggested if the derivative of $\displaystyle \displaystyle z^2 + 1$, i.e. $\displaystyle \displaystyle 2z$ is a factor.
The integral of $\displaystyle \displaystyle \frac{1}{1 + x^2}$ is $\displaystyle \displaystyle \arctan{x}$.
You see there's me being stupid again thinking d/dz z^2 = 2, of course it isn't! my brain sometimes, seriously :S
Yeah I get that I just miscalculated what I was thinking of. Thanks for putting me back in my place lol.
so y(t) = tan(t+c) - t, that is the final solution? that's what I have.