## FibonacciChick666 Group Title 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) 11 months ago 11 months ago

1. abb0t Group Title

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

2. FibonacciChick666 Group Title

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

3. FibonacciChick666 Group Title

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

4. FibonacciChick666 Group Title

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

5. FibonacciChick666 Group Title

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

6. dumbcow Group Title

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

7. FibonacciChick666 Group Title

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

8. dumbcow Group Title

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

9. FibonacciChick666 Group Title

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

10. abb0t Group Title