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s3a
 4 years ago
(Exact Differential Equation)
In attempting this question:
http://f.imgtmp.com/JYceq.jpg
, I did this:
http://f.imgtmp.com/New69.jpg
. I feel I should be right but Wolfram Alpha gives an entirely different answer (which doesn't seem compatible as an answer for the question):
http://www.wolframalpha.com/input/?i=(x^3+%2B+4y)+dx+%2B+(4x+%2B+4y)+dy+%3D+0.
Could someone please help me find what's wrong?
s3a
 4 years ago
(Exact Differential Equation) In attempting this question: http://f.imgtmp.com/JYceq.jpg , I did this: http://f.imgtmp.com/New69.jpg . I feel I should be right but Wolfram Alpha gives an entirely different answer (which doesn't seem compatible as an answer for the question): http://www.wolframalpha.com/input/?i=(x^3+%2B+4y)+dx+%2B+(4x+%2B+4y)+dy+%3D+0. Could someone please help me find what's wrong?

This Question is Closed

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.1I couldn't read your writing, but here how I would it. \[(x^3 + 4y) dx + (4x + 4y) dy = 0\] This differential equation is obviously exact, so there exists a solution \(f(x,y)=c\) such that \(f_x=x^3+4y\) and \(f_y=4x+4y\). Then, by integrating the first equation with respect to x we get \(\large f(x,y)=\frac{x^4}{4}+4xy+g(y)\) (*) Now, differentiating (*) with respect to y gives: \[f_y=4x+g'(y)=4x+4y \implies g'(y)=4y \implies g(y)=2y^2+k\] Hence the solution is\(\frac{x^4}{4}+4xy+2y^2=c.\) This solution is good enough for me, but if you want to find a similar form to what wolfarm is giving then you can solve this last equation as by using the quadratic formula in y.

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.0Wolfram solution also solves for the implicit function problem, but this is not part of the differential equation problem...
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