Using either Method 1: Standard form[y' +p(t)y = q(t)] or Method 2: Ygeneral = Yh(homogeneos) + Yp(particular)

1. y' = (1 / (t+1) )y + 3t^2 + 3t ; y(0) = 5

2. y' = −2ty + 3e^(-t^2) ; y(0) = 4

god I didnt remember these part about it.
I wish I didnt have to retry these class again (failed integral.. thus the whole chunk of differential + integral -.-)

These can take a while to work out, so I'll outline the steps for you for #1. (#2 is similar)

First, rearrange the equation to standard form: y' - (1 / (t+1))y = 3t^2 + 3t

Then, find the integrating factor (call it mu(t)) of the form mu(t) = e^integral(p(t)dt). In case you're not aware, p(t) = - 1 / (t+1) in this case.

Once you have mu(t), multiply both sides of the equation by it. This will give you mu(t)(y' - (1 / (t+1))y = mu(t)(3t^2 + 3t).

Simplify the left hand side: mu(t)(y' - (1 / (t+1))y = d/dt (mu(t)y). This gives you d/dt (mu(t)y) = mu(t)(3t^2 + 3t).

Integrate both sides and you'll have mu(t)y = integral(mu(t)(3t^2 + 3t)dt). Evaluate the integral on the right hand side (don't forget + C as it's indefinite).

Once you have this, multiply both sides by mu^-1(t) to isolate y(t) by itself. Plug in the initial conditions and you'll have your solution.