## 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?

1. Mr.Math

I 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.

2. cristiann

Wolfram solution also solves for the implicit function problem, but this is not part of the differential equation problem...