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 September 30th, 2012, 10:40 AM #1 Newbie   Joined: Sep 2012 Posts: 14 Thanks: 0 Exact differential equation Hey guys, I've been trying to solve a differential equation, I checked 20 times but it's not the right answer.. It's dy/dx=(3*x^3+y)/(-x-4*y^3) wich equals (3*x^3+y)dx+(x+4*y^3)dy=0 So My=1 and Nx=1, so the equation is exact Then I do the integer of 3*x^3+y, which gives 3/4*x^4+x*y+(a function of y) I do all the steps and I find the function of y, which is y^4 so my answer is 3/4*x^4+x*y+y^4 but it says it's not the right answer. Could anyone help me please?
 September 30th, 2012, 10:57 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Exact differential equation You have correctly expressed the ODE in differential form: $$$3x^3+y$$\,dx+$$x+4y^3$$\,dy=0$ You have correctly determined the equation is exact. So, we set: $F(x,y)=\int 3x^3+y\,dx+g(y)$ $F(x,y)=\frac{3}{4}x^4+xy+g(y)$ Now differentiate with respect to y: $x+4y^3=x+g#39;(y)$ $g'(y)=4y^3$ $g(y)=y^4$ Hence, the solution is given implicitly by: $\frac{3}{4}x^4+xy+y^4=C$ It appears the only thing you have left out is the constant.
 September 30th, 2012, 11:14 AM #3 Newbie   Joined: Sep 2012 Posts: 14 Thanks: 0 Re: Exact differential equation That's what I thought but I keep entering this answer and it says that it's wrong... maybe there's a problem on the site.. Thanks a lot!
 October 1st, 2012, 07:02 PM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Exact differential equation The way I would have done this problem is this: to say that $(3x^3+ y)dx+ (x+ 4y^3)dy= 0$ is an exact equation means that there exist a function F(x,y) such that $dF= F_xdx+ F_ydy= (3x^3+ y)dx+ (x+ 4y^3)dy$. So we must have $F_x= 3x^3+ y$ and differentiating with respect to x treats y as a constant, integrating, treating y as a constant, $F(x, y)= (3/4)x^4+ xy+ g(y)$ where, because we are treating y as a constant, the "constant of integration" may be a function of y, g(y). Differentiating that with respect to y, $F_y= x+ g#39;(y)= x+ 4y^3$. That is, we must have $g'(y)= 4y^3$ so that $g(y)= y^4+ C$ where, because g is a function of y only, C really is a constant. That is $F(x,y)= (3/4)x^4+ xy+ y^4+ C$. Because the original differential equation was "dF= 0", F is a constant: $F(x,y)= (3/4)x^4+ xy+ y^4+ C= C#39;$ and we can combine the two constant to get $(3/4)x^4+ xy+ y^4= C$.

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