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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

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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?
 
416
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the eqn is (x-1)dx + (y+1)dy=0 , y(1)=0
just only this
 

berkeman

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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?
 

berkeman

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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.
 
416
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Are the following differential equations exact? Solve the IVP:

(x-1)dx + (y+1)dy=0 , y(1)=0
 

berkeman

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Sorry, I'm not familiar with that form of differential equation. I'll see if I can find somebody to help.
 
416
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okok........thanx
 

HallsofIvy

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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.
 
416
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thanx............
 

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