Solving Initial Value Problem: Finding c

In summary, the conversation was about solving an initial value problem and finding the value of c. The equation (x-1)dx + (y+1)dy=0 with y(1)=0 was given, and it was mentioned that the equation is separable and can be integrated directly to find the value of c. Further clarification was requested about the problem, but the conversation ended before a resolution was reached.
  • #1
teng125
416
0
i try to solce the initial value problem question and got f(x,y) = x^2 -x +y^2 +y +c and y(0)=2 is given

how do i find c??
is it just make the eqn x^2 -x +y^2 +y +c = 0 and subs y=2 and x=0??

pls help

thanx
 
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  • #2
I think you need more info. To find c, you would need to know f(0,0), x(0) and y(0). Are those available?
 
  • #3
the eqn is (x-1)dx + (y+1)dy=0 , y(1)=0
just only this
 
  • #4
teng125 said:
the eqn is (x-1)dx + (y+1)dy=0 , y(1)=0
just only this
I don't understand the problem. Could you please post the full text of the problem? Is it an integration or differentiation or what? Are you taking partial derivatives?
 
  • #5
And what do you mean by both y(1)=0 and y(0)=2? What is the argument to y? Time? Too many things are getting mixed up here.
 
  • #6
Are the following differential equations exact? Solve the IVP:

(x-1)dx + (y+1)dy=0 , y(1)=0
 
  • #7
Sorry, I'm not familiar with that form of differential equation. I'll see if I can find somebody to help.
 
  • #8
okok...thanx
 
  • #9
teng125 said:
Are the following differential equations exact? Solve the IVP:

(x-1)dx + (y+1)dy=0 , y(1)=0

Of course it's exact: [itex]\frac{\partial (x-1)}{\partial y}= \frac{\partial (y+1)}{\partial x}= 0[/itex]. In fact, it's separable.
(x-1)dx+ (y+1)dy= 0 can be integrated directly to get
[tex]\frac{1}{2}x^2- x+ \frac{1}{2}y^2+ y+ C= 0[/itex]
Now put x= 1, y= 0 to find C.
 
  • #10
thanx...
 

1. How do I find the value of c in an initial value problem?

To find the value of c in an initial value problem, you will need to solve the differential equation using the given initial conditions. This will give you an equation with c as the only unknown variable. You can then solve for c using algebraic methods.

2. What is the importance of finding the value of c in an initial value problem?

The value of c in an initial value problem is important because it represents the constant term in the solution to the differential equation. It is the specific value that makes the solution satisfy the given initial conditions.

3. Can there be multiple values of c that satisfy an initial value problem?

Yes, it is possible for there to be multiple values of c that satisfy an initial value problem. This can happen when the initial conditions are not enough to uniquely determine the value of c, or when the equation has multiple solutions.

4. What happens if I cannot find the value of c in an initial value problem?

If you cannot find the value of c in an initial value problem, it may mean that the equation does not have a solution or that the initial conditions are not compatible with the equation. In this case, you may need to re-examine your calculations or adjust the initial conditions.

5. Are there any techniques or methods that can help me find the value of c in an initial value problem?

Yes, there are several techniques and methods that can help you find the value of c in an initial value problem. These include separation of variables, substitution, and using an integrating factor. It is important to choose the most appropriate method for the specific equation and initial conditions given.

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