Solving Initial Value Problem: Finding c

[teng125]

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

 

[berkeman]

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?

 

[teng125]

the eqn is (x-1)dx + (y+1)dy=0 , y(1)=0
just only this

 

[berkeman]

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?

 

[berkeman]

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.

 

[teng125]

Are the following differential equations exact? Solve the IVP:

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

 

[berkeman]

Sorry, I'm not familiar with that form of differential equation. I'll see if I can find somebody to help.

 

[teng125]

okok...thanx

 

[HallsofIvy]

Are the following differential equations exact? Solve the IVP:

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

Of course it's exact: $\frac{\partial (x-1)}{\partial y}= \frac{\partial (y+1)}{\partial x}= 0$. 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]<br /> Now put x= 1, y= 0 to find C.[/tex]

 

[teng125]

thanx...