Integrating Functions with Variables: Solving Differential Equations

Click For Summary
The discussion revolves around solving the differential equation f'(x) = x + 1 - yx - y, with a focus on integrating functions that include variables. The user initially integrated both sides, leading to f(x) = x^2/2 + x - integral(yx) - integral(y) + c, but struggled with the presence of y in the function. It was clarified that the equation can be treated as a separable ordinary differential equation (ODE) by rewriting it as y' = (x + 1) - y(x + 1), allowing for the separation of variables. The confusion stemmed from interpreting f(x) as a function of y, but it was noted that f(x) and y can be related through dy/dx. Ultimately, the solution involves recognizing the separable nature of the equation to proceed with integration.
CalculusHelp1
Messages
22
Reaction score
0

Homework Statement



f'(x)= x+1-yx-y

Solve for f(x)


Homework Equations



definitions of integrals and derivatives


The Attempt at a Solution



Okay, so I integrated both sides.

This gives f(x) = x^2/2 + x -integral(yx) - integral(y) +c

The problem I'm having is how do you integrate when there is a y in the function? I'm guessing this has something to do with the definition, i.e. f(x)=y=integral(f'(x)), I'm just not sure how to manipulate the equation. Does anyone know the trick to this?
 
Physics news on Phys.org
So you've got the equation y'=x+1-yx-y. This is a separable ODE, you can write it as y'=(x+1)-y(x+1). Thus y'/(1-y)=x+1.
Can you solve it now?
 
Thanks again micromass
 
micromass said:
So you've got the equation y'=x+1-yx-y. This is a separable ODE, you can write it as y'=(x+1)-y(x+1). Thus y'/(1-y)=x+1.
Can you solve it now?

ummm but he stated f(x), not a function of y. Unless I'm mistaken
 
Sorry i was given dy/dx and was asked to solve for y, i just replaced with with f(x) and dy/dx with f'(x). The solution was to separate the variables, I forgot all about it.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
7
Views
2K