SOLUTION: Question (a) Determine whether the differential equation ydx - xdy =0 is exact: (b) If not, convert it into an exact differential equation_ (c) Solve this differential equat

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Question 1199306: Question
(a) Determine whether the differential equation
ydx - xdy =0 is exact:
(b) If not, convert it into an exact differential equation_ (c) Solve this differential equation.

Found 3 solutions by Edwin McCravy, ikleyn, mccravyedwin:
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!



That is easily solvable by separating the variables.  Its solution is y = Cx

But your professor wants you to tackle it as either an exact differential 
equation or one solvable by using an integrating factor to convert it to
an exact differential equation. 

First we check to see if the differential equation is exact:

We write it in the form 



So M(x,y) = y,   N(x,y) = -x

We form the partial derivatives:

, 

Those are not equal, so the differential equation is not exact.

So we can try one of these integrating factors to multiply through by to
see if it comes out to be an exact differential equation:



Trying the first one:



So we multiply the original differential equation



 through by 



Let's see if that is exact.

Taking the partial of  with respect to y givss 

Taking the partial of  with respect to x 
gives 

They are not equal so this integrating factor did not work.  There is no
use to try the other one because the equation is symmetrical in x and y,
and it would be the same with the x's and y's swapped.

So we can't convert the given differential equation to an exact one by
the usual method.  That's because the method assumes it's possible to
find an integrating factor in terms of one variable only.  So it's not
possible.  So we can't do what your professor asked you to do.

So let's just solve it by separating the variables:



I like my dy to be at the first:





Divide both sides by 















Edwin

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

            Below is my note to the solution by  Edwin.


I  fully agree on how Edwin deduced equation 

          (*)


But what follows, should be corrected, because 

     is not ln(y):  it is ln(|y|);     is not ln(x):  it is ln(|x|)

                            with y =/= 0;                            with x =/= 0.


Therefore, from (*) we have

    ln(|y|) = ln(|x|) + ln(C),  x =/= 0;  y =/= 0

    y = Cx,  with  x =/= 0,;  y =/= 0;  C may have any sign, positive or negative, except C = 0.



////////////////


It is with regret I see that Edwin (@mccravyedwin) ascribes to me what I did not say and did not write.


                        Edwin,  it is a prohibited way to make a discussion.

                            I said what I said,  and what  I  said was right.


Happy New Year !



Answer by mccravyedwin(407)   (Show Source): You can put this solution on YOUR website!
Ikleyn claims that one must never write "ln(x)" but only "ln(|x|)".  But most any
teacher would write "ln(x)" and "logb(x)".  I suppose Ikleyn would also
say that in real numbers one must never write √(x) but only √(|x|).  However,
domains are well-known for these functions in real numbers.  It is not necessary
to complicate the notation.

To prove deMoivre's theorem students are often taught Euler's equation:

 which implies  or 

In complex analysis, an undergraduate mathematics course, logarithms of negative
numbers are well defined as complex numbers.  

Edwin

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