General solution of diffential equation

delsoo
Messages
97
Reaction score
0

Homework Statement



i can't reduce the eq to y= (X+1)/(x+c ) , which part is wrong?

Homework Equations





The Attempt at a Solution

 

Attachments

  • DSC_0143~5[1].jpg
    DSC_0143~5[1].jpg
    23.7 KB · Views: 398
  • DSC_0144~4[1].jpg
    DSC_0144~4[1].jpg
    22.7 KB · Views: 385
Physics news on Phys.org
You already have it.
y=\frac{c(1+x)}{1+c(1+x)}
Divide numerator and denominator by c:
y=\frac{(1+x)}{(1+1/c)+x}
Let C = 1 + 1/c
y=\frac{(1+x)}{C+x}
Chet
 
can we straight sub C =1 + 1/c ? the c is the question not the original c (constant ) from the integrate( (dx/X+1)) ? dx ?
 
delsoo said:
can we straight sub C =1 + 1/c ? the c is the question not the original c (constant ) from the integrate( (dx/X+1)) ? dx ?

Of course.

Chet
 
the c is the question the original c (constant ) from the integral ( (dx/X+1)) ? dx ? am i right?
 
delsoo said:
the c is the question the original c (constant ) from the integral ( (dx/X+1)) ? dx ? am i right?
In your original solution, the choice of adding ln c as the constant of integration was not unique. You could just as easily have added a different constant, say c' (i.e., without the natural log), without affecting the correctness of the solution.

Chet
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top