Question

How do I solve a differential equation if the equation becomes undefined when I plug in the initial value?

Answer 1

heyllo joseph hru?

Answer 2

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

Answer 3

dx/dy = (y-1)^2

Answer 4

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

Answer 5

y(0) = 1

Answer 6

C = 0

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

excuse me! dy/dx =...

Answer 11

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
\[x = (y-1)^3/3\]

Answer 12

+C

Answer 13

DON'T FORGET THE +C

Answer 14

k

Answer 15

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

Answer 16

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.