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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Show the equation :D If it's undefined at the initial value, then you probably integrated incorrectly :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = 1/3(y1)^3 + C What are the initial conditions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you show your work in getting the solution to the differential equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y(0) = 1 x = 0 y = 1 0 = 1/3(11)^3 +C 0 = 0 +C C = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0excuse me! dy/dx =...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh:\[dx/dy = (y1)^2\] \[\int\limits_{0}^{x}x dx = \int\limits_{0}^{y} (y1)^2 dy\] Take the antiderivative (in this case, it works) :D \[x = (y1)^3/3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0excuse me, i made a mistake, the dy and dx are flipped from what I wrote. Is it OK?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, then the equation is completely different: \[dy/dx = (y1)^2\] Limits using initial conditions \[\int\limits_{1}^{y} dy/(y1)^2 = \int\limits_{0}^{x} xdx\] \[\int\limits_{1}^{y} (y1)^{2} dy = x\] Antidifferentiate again:\[(y1)^{1}   (0)^{1} = x \] No solution for this diff eq.
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