ODE with integrating factor NEED HELP
1. The problem statement, all variables and given/known data
[tex] (2x+y^2) dx +4xy dy=0,y(1)=1 [/tex] 2. Relevant equations 3. The attempt at a solution I'm having trouble finding the correct integrating factor, been playing with it for an hour and have made NO progress so need help. [tex] \delta P/\delta y=2y [/tex] [tex] \delta Q/\delta x=4y [/tex] so its not exact and I need to find an integrating factor. [tex] R=1/Q( \delta P/\delta y\delta Q/\delta)=1/2x [/tex] [tex] F(x)= exp \int R(x) dx [/tex] [tex] \int 1/(2x) = ln(abs(x))/2 [/tex] This integrating factor isn't going to make my ode exact though and I can't figure out what I have stuffed up??? Thanks for any help. 
Re: ODE with integrating factor NEED HELP
Quote:
[tex]R(x) = e^{\int\frac 1{2x}dx}= e^{\frac 1 2 \int\frac 1 x}\, dx = e^{\frac 1 2 \ln x} = e^{\ln(x^{\frac 1 2})} = x^{\frac 1 2}[/tex] Also, a TeX note, you can use the partial symbol [tex]\partial Q/\partial x[/tex] but subscripts are even easier: Q_{x} or [itex]Q_x[/itex]. 
Re: ODE with integrating factor NEED HELP
Thanks LCKurtz I now have a general and particular solution, would you mind checking my results:
[tex] u(x,y)=4x^{3/2}+2*\sqrt(x)*y^2+y=c [/tex] [tex] c=13/3 [/tex] [tex] 13/3=4x^{3/2}+2*\sqrt(x)*y^2+y [/tex] THANKS ps should be 4x to the power of 3/2 , doesn't seem to like my tex. 
Re: ODE with integrating factor NEED HELP
Quote:

Re: ODE with integrating factor NEED HELP
I don't know what you mean by Check that du = Pdx + Qdy, du as in the differential of
[tex] 4x^{3/2}+2*\sqrt(x)*y^2+y [/tex] also should [tex] 4*1^{3/2}+2*\sqrt(1)*1^2+1 [/tex] = 13/3? because it =s 7?? am I doing something wrong with my checking or is my solution wrong? Thanks 
Re: ODE with integrating factor NEED HELP
The whole point of finding an integrating factor is converting your differential equation into an exact one. What LCKurtz is suggesting is that you use the partial derivative test to see if it is indeed exact.

Re: ODE with integrating factor NEED HELP
Quote:
Quote:
Quote:
Quote:
[tex](*)\ (2x^{\frac 1 2}+ 2x^{\frac 1 2}y^2)dx+4x^{\frac 1 2}y dy = 0[/tex] Presumably you have checked that this equation is exact and you have a proposed solution u(x,y) written above. The way to check whether your solution is correct is to calculate du = u_{x}dx + u_{y}dy and see if it agrees with the exact equation (*) above. Once you have that you need to check that x = 1, y = 1 works in your final answer. 
Re: ODE with integrating factor NEED HELP
Decided to practice a bit on this one. Here's what I've got:
1. Orig. DE is exact with the integ. factor: [tex]\mu=1/\sqrt{x}[/tex] because [tex]\frac{\partial(\frac{2x+y^2}{\sqrt{x}})}{\partial y}=\frac{2y}{\sqrt{x}}=\frac{\partial(4\sqrt{x}y)}{\partial x}\Longleftrightarrow\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}[/tex] 2. To solve for u, we have: [tex]u(x,y)=\int\frac{2x+y^2}{\sqrt{x}}dx+g(y)=\frac{4x^{3/2}}{3}+2\sqrt{x}y^2+g(y)[/tex] 3. To find g(y), we take the partial derivative of the newly found function u(x,y) in (2) with respect to y and set it equal to Q, that is: [tex]\frac{\partial u}{\partial y}=4\sqrt{x}y+\acute{g(y)}\Longrightarrow 4\sqrt{x}y+\acute{g(y)}=4\sqrt{x}y \Longrightarrow \acute{g(y)}=0[/tex] 4. So, after integrating the result in (3), we find that: [tex]g(y)=c_{1}[/tex] Thus, the overal function u(x,y) takes the form: [tex]u(x,y)=\frac{4x^{3/2}}{3}+2\sqrt{x}y^2+c_{1}=c_{2}[/tex] or, combining the constants, we have: [tex]\frac{4x^{3/2}}{3}+2\sqrt{x}y^2=c[/tex] 5. If y(1)=1, the constant c works out to be c=10/3, so that the unique solution becomes: [tex]\frac{4x^{3/2}}{3}+2\sqrt{x}y^2=\frac{10}{3}[/tex] 6. A simple check shows that: [tex]du=u_{x}dx+u_{y}dy=(\frac{4}{3}\cdot\frac{3}{2}x^{1/2}+2\frac{1}{2}x^{1/2}y^2)dx+(2\cdot2\sqrt{x}y)dy=(\frac{2x+y^2}{\sqrt{x}})dx+(4\sqrt{x}y)d y[/tex] which is the original DE multiplied by the integrating factor. 7. I learned some simple TEX commands o:) 
All times are GMT 5. The time now is 10:58 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums