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1+x^2 dy/dx = 1+y^2, differential equation! a little Help here! 
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#1
Aug3110, 04:45 PM

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1. The problem statement, all variables and given/known data
1+x^2 dy/dx = 1+y^2 2. Relevant equations 3. The attempt at a solution if I clean this up a little, I would get 1/ (1+x^2) dx = 1/ (1+y^2) dy correct? since the integral of 1+x^2 is arctanX, I get arctanX + C = arctanY + C. And I don't know what to do from here. Please help! 


#2
Aug3110, 05:23 PM

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P: 1,583

You have:
[tex] \tan^{1}y=\tan^{1}x+\tan^{1}(d) [/tex] Take tan of both sides and use the double angle formula for tan. 


#3
Sep110, 08:23 AM

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Now, if you like you can write arctan(y)= arctan(x)+ (C1 C2) and since C1 and C2 are unknown constants, C1 C2 can be any constant. Just call it "C": C= C1 C2. Then you have arctan(y)= arctan(x)+ C. If you want to solve for y, just take the tangent of both sides: y= tan(arctan(x)+ C) Caution this is NOT the same as tan(arctan(x))+ arctan(C)! What is true is that [tex]tan(A+ B)= \frac{tan(A)+ tan(B)}{1+ tan(A)tan(B)}[/tex] with A= arctan(x) and B= C, tan(A)= x so [tex]y= tan(arctan(x)+ C)= \frac{x+ tan(C)}{1+ tan(C)x}[/tex] and, again, since C is an arbitrary constant and the range of tangent is all real numbers, tan(C) is an arbitrary constant. letting tan(C)= C', [tex]y= \frac{x+ C'}{1+ C'x}[/tex] 


#4
Sep110, 11:47 PM

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1+x^2 dy/dx = 1+y^2, differential equation! a little Help here!
Sorry, HallsofIvy, it is not true. The correct formula is: [tex]tan(A+ B)= \frac{tan(A)+ tan(B)}{1 tan(A)tan(B)}[/tex] By the way, the original equation 1+x^2 dy/dx = 1+y^2 is equivalent to x^2 dy/dx = y^2 with the solution [tex]y=\frac{x}{1+cx}[/tex]: ehild 


#5
Sep210, 12:02 AM

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The joys of sloppy use of parentheses.
It looks like the OP meant to have a pair of parentheses around 1+x^{2}, but I suppose we should let him clarify rather than assuming one way or another. 


#6
Sep210, 12:36 AM

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ehild 


#7
Sep210, 07:17 AM

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By the way, the original equation 


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