## anonymous 5 years ago How do I solve a differential equation if the equation becomes undefined when I plug in the initial value?

1. anonymous

heyllo joseph hru?

2. anonymous

Show the equation :D If it's undefined at the initial value, then you probably integrated incorrectly :D

3. anonymous

dx/dy = (y-1)^2

4. anonymous

x = 1/3(y-1)^3 + C What are the initial conditions?

5. anonymous

y(0) = 1

6. anonymous

C = 0

7. anonymous

x = 1/3 (y-1)^3

8. anonymous

can you show your work in getting the solution to the differential equation?

9. anonymous

y(0) = 1 x = 0 y = 1 0 = 1/3(1-1)^3 +C 0 = 0 +C C = 0

10. anonymous

excuse me! dy/dx =...

11. anonymous

Oh:$dx/dy = (y-1)^2$ $\int\limits_{0}^{x}x dx = \int\limits_{0}^{y} (y-1)^2 dy$ Take the antiderivative (in this case, it works) :D $x = (y-1)^3/3$

12. anonymous

+C

13. anonymous

DON'T FORGET THE +C

14. anonymous

k

15. anonymous

excuse me, i made a mistake, the dy and dx are flipped from what I wrote. Is it OK?

16. anonymous

No, then the equation is completely different: $dy/dx = (y-1)^2$ Limits using initial conditions $\int\limits_{1}^{y} dy/(y-1)^2 = \int\limits_{0}^{x} xdx$ $\int\limits_{1}^{y} (y-1)^{-2} dy = x$ Antidifferentiate again:$-(y-1)^{-1} - - (0)^{-1} = x$ No solution for this diff eq.