Hey, Quick help Please: Transform the given IVP into an equivalent problem with the initial pt. at the origin. $\frac{dy}{dt}=t^2+y^2$ (I know it's easy... I'm just having a brain fart)

Question

Hey, Quick help Please: Transform the given IVP into an equivalent problem with the initial pt. at the origin.  $\frac{dy}{dt}=t^2+y^2$  (I know it's easy... I'm just having a brain fart)

abb0t · Answer

put it in the form $y'-t^2=y^2$

fibonaccichick666 · Answer

whoops forgot the IV $y(1)=2$

fibonaccichick666 · Answer

I need a little more please @abb0t ... It's not my ODE type of day

fibonaccichick666 · Answer

Hey @hero could you take a look at this for me please, it's just not clicking  at the moment

fibonaccichick666 · Answer

@amistre64, @mathstudent, or @satellite  Do you mind taking a look at this for me please?  You come  highly recommended

dumbcow · Answer

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

fibonaccichick666 · Answer

oh.... that was easy thanks... I had y'= 5t and I'm like this is sooo wrong ok ty again

dumbcow · Answer

yw i hope you dont have to solve this one...its ugly no elementary solution

fibonaccichick666 · Answer

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

abb0t · Answer

I see someone helped you already. I had to go sorry.