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

[fredrick08]

and for g(y)/g'(y)=a Integral of all of it is G(y)/g(y)=ay=>g(y)=G(y)/(a*y)??

 

[Cyosis]

Are you just randomly guessing or? We want to SOLVE f(x). We want to find all expressions for f(x) that satisfy this differential equation. Have you never solved a single variable differential equation before?

You're now saying that the integral of a fraction is the integral of the numerator divided by the integral of the denominator. This is wrong.

Example:
$$\int \frac{1}{x}dx=\log x \neq \frac{\int 1 dx}{\int x dx}=\frac{2}{x}$$

Hint: make a substitution u=f(x)

 

[fredrick08]

ok sorry... i didnt realize how stupid that was... integral of all it is constant*ln(x)=ax...

 

[Cyosis]

That is almost correct. It should be $\ln (f(x))=ax+cst$. Do you see why? If not use the hint in my previous post, which I edited in just before your reply.

 

[fredrick08]

oh yes ln(f(x))=ax+c... so then for g... (g(y)/2)y^2=ay+c?

 

[Cyosis]

The integral we want to calculate is:

$$\int \frac{f'(x)}{f(x)}dx$$

Use the substitution u=f(x). You do know how to integrate by using substitutions right?

 

[fredrick08]

its something liek that... because no matter what number you put infront of g... eg 3y^2, then g'=6y integral of g/g'= ((3*2)/2)y^2

ok ignore this

 

[Cyosis]

Can you please just evaluate the integral and stop guessing? And if you guess don't guess things that are so obviously wrong, g/g'=constant. Yet in your example it's suddenly a function of y?

Evaluate the integral with the substitution I have given you!

 

[fredrick08]

ok then so int(f'/f)=ln(f(x))=ax+c... yes i can intergrate by substituion... but what am i intergating? ln(u)=ax+c

 

[Cyosis]

You're asking me what we are integrating yet you give the answer of the integration. This means to me that you did not evaluate the integration. Show me the steps leading up to your answer.

 

[fredrick08]

anrt i meant to solve for u? therefore u=f(x)=e^(ax+c)?

 

[fredrick08]

ok we are integrating f'/f=a so the answer would be ln(f(x))=ax+c (as u said), then you said something about substitution...

 

[Cyosis]

Fredrick, if you just ignore what I ask of you we are going nowhere. You do not need to solve for u. I want you to evaluate the integral, in other words simply solve the differential equation f'(x)/f(x)=a.

Can you do this or are we going to continue to beat around the bush?

ok we are integrating f'/f=a so the answer would be ln(f(x))=ax+c (as u said), then you said something about substitution...

How did you get that answer, that is what I want to know.

Understanding how you arrive at that answer will allow you to understand how to solve the entire system.

 

[fredrick08]

i don't know that is wat u told me... but if use f(x)=3x^2 then int of f'(x)/f(x) dx would equal 2 ln(x).. this is what i am confused about as don't understand what you mean by solve the differential equation... as i am unsure how to do so, without a function...

 

[Cyosis]

To be able to solve partial differential equations you must first be able to solve single variable differential equations. You must also be able to integrate by substitutions. Have you followed these courses?

I have given you a substitution that allows you to solve the integral, u=f(x). What is du?

 

[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.

$$\begin{align*} \int \frac{f'(x)}{f(x)}dx & = \int a dx \\ \int \frac{1}{u}du & =ax+c \\ \log u & =ax+c \\ \log f(x) & =ax+c \\ f(x) & =e^{ax+c} =Ae^{ax} \end{align*}$$

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.