Partial Diff Qn: Can u(x,y) be a Product of fx and fy?

[fredrick08]

ok so you want me to use substitution, with u=f(x) on

int(f'(x)/f(x))dx then int(u'/u)du??

 

[Cyosis]

No that is wrong. Let's try something simple first if u=x^2 then du=?

 

[fredrick08]

ok i think more about it... its really simple, just finding it hard to remember...

 

[fredrick08]

ok lol I am dumb... int(f'/f)dx=int((1/u)(du/dx)du=int(1/u)du=ln(u)=ln(f(x))

 

[fredrick08]

ok so that equals int(a)dx=ax+c rite? but then do i solve for f(x), so f(x)=e^(ax+c)?

 

[Cyosis]

Okay I will do this example for you. I however strongly advice you to review basic calculus topics like single variable linear differential equations and basic integration techniques.

We want to solve f'(x)/f(x)=a. To do this we integrating both sides of the equation then substitute u=f(x), therefore du=f'(x)dx.

<br /> \begin{align*}<br /> \int \frac{f&#039;(x)}{f(x)}dx &amp; = \int a dx<br /> \\<br /> \int \frac{1}{u}du &amp; =ax+c<br /> \\<br /> \log u &amp; =ax+c<br /> \\<br /> \log f(x) &amp; =ax+c<br /> \\<br /> f(x) &amp; =e^{ax+c} =Ae^{ax}<br /> \end{align*}<br />

Edit: I see you got it now, good. Now do the same for g(x).

 

[fredrick08]

ok sorry for g(x)...

int(g/g')dx=int(u*(dx/du)du=int(u)dx=ux=xf(x)??

=>f(x)=(ax+c)/x??

 

[Cyosis]

Nope that is wrong. Try to invert the function first so that the fraction looks the same as the one for f. Secondly g is a function with variable y not x.

 

[fredrick08]

no that's not rite... omg int(u(dy/du)dy=int(u)dy^2/du? ok no idea what i am doing here

 

[Cyosis]

Invert the equation and substitute u=g(x).

 

[fredrick08]

can i just flip it?? then it will be the same as f... =ln(g(y))?

 

[Cyosis]

Yes you can flip one side, but then you also need to flip the other side.

 

[fredrick08]

ok so ln(g(y))=(ax+c)^-1??

 

[Cyosis]

No you are solving the differential equation g(y)/g'(y)=a , then g'(y)/g(y)=?

 

[fredrick08]

1/a? so ln(g(y))=(y/a)+c?? =>g(y)=e^((y/a)+c)

 

[Cyosis]

No. How can the integral of f'/f have a different value than g'/g. If g(y)/g'(y)=a then g'(y)/g(y)=1/a. Now integrate.

 

[fredrick08]

yes isn't that what i did? int(1/a)dy=(y/a)+c right?

 

[Cyosis]

Ugh sorry it's 4 am here I am getting sleepy! Yes you're right so what is the the total solution to the partial differential equation you began with?

 

[fredrick08]

u(x,y)=f(x)g(y)=Ae^(ax)*Be^(y/a)... but I am not sure... if i differentiate that, i don't thik it works out.

 

[Cyosis]

It's correct, just differentiate it and you will see it will turn out correctly.

 

[fredrick08]

ok omg thankyou very much... very much appreciated = ) have a good sleep

 

[Cyosis]

You're welcome, thanks and good night!