## anonymous 2 years ago *Solve the differential equation dy/dx=x/16y. *Find the equation of the solution through the point (x,y)=(−4,1). *Find the equation of the solution through the point (x,y)=(0,−4). Your answer should be of the form y=f(x)

1. anonymous

I already solved the first part. C=8y^2 -(x^2/2)

2. myininaya

ok replace your x and y with -4 and 1 respectively to find the C so that your graph for your equation will go through that point

3. anonymous

so y=sqrt((x^2/16) + c) and i plugged in (-4,1) and got y=sqrt((x^2/16) is that correct?

4. anonymous

c= 0 for the second question?

5. myininaya

You have $C=8y^2-\frac{x^2}{2}$ Replace x with -4 and y with 1 like this: $C=8(1)^2-\frac{(-4)^2}{2}$ Is this what you did to solve for C?

6. myininaya

$C=8-\frac{16}{2}=8-8=0$

7. anonymous

yes that's what i did

8. myininaya

I can't do arithmetic somethings lol But you have $0=8y^2-\frac{x^2}{2}$

9. myininaya

that is the equation satisfying the point (-4,1)

10. anonymous

hmm it the system doesn't seem to accept it.. it says it isn't a linear equation

11. anonymous

doesn't that mean i have to solve for y=...

12. myininaya

Your differential equation was: $\frac{dy}{dx}=\frac{x}{16y} \text{ \right ?}$

13. anonymous

yup

14. myininaya

Ok solve for y make sure you keep in mind we want to go through the point (-4,1)

15. anonymous

ok so y=x/4

16. anonymous

but it still doesn't work??

17. anonymous

oh wait y=-x/4 works!!

18. myininaya

We wanted y=-x/4 That is why I said to keep in mine we want it going through (-4,1) :)

19. myininaya

mind*

20. myininaya

The equation y=x/4 doesn't satisfy the point (-4,1) but y=-x/4 does

21. myininaya

Like 1=-(-4)/4

22. anonymous

ohh i get it now

23. anonymous

thanks a lot!!

24. anonymous

i'm going to do the last part now

25. anonymous

c=128 for the last part right?

26. anonymous

oh i got the last part too!!

27. anonymous

thanks a lot again!!!

28. myininaya

yes C=128 for the last one