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put it in the form \(y'-t^2=y^2\)
whoops forgot the IV \(y(1)=2\)
I need a little more please @abb0t ... It's not my ODE type of day
oh oops i was trying to solve the ODE...so you just want to change ODE to reflect the IV at (0,0)
plug in original IV (1,2) to determine dy/dt
dy/dt = 1^2 + 2^2 = 5
now plug in (0,0) and keep dy/dt = 5
5 = 0^2 +0^2 + C
5 = C
dy/dt = t^2 +y^2 +5
oh.... that was easy thanks... I had y'= 5t and I'm like this is sooo wrong ok ty again
yw i hope you dont have to solve this one...its ugly no elementary solution
yea I know... I just have a massive headache and screwed this up like 6 times so I was sick of it lol but nah just change to IV =0 but I was thinking integrating factor maybe
I see someone helped you already. I had to go sorry.