1+x^2 dy/dx = 1+y^2, differential equation! a little Help here!


by shseo0315
Tags: differential, dy or dx, equation
shseo0315
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#1
Aug31-10, 04:45 PM
P: 19
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!
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hunt_mat
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#2
Aug31-10, 05:23 PM
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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.
HallsofIvy
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Sep1-10, 08:23 AM
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Quote Quote by shseo0315 View Post
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!
First, you don't need both "C"s. And you especially should not write the same symbol, "C", for both- that makes it look like they will cancel. Since the constant of integration may not be the same on both sides, you shold write "arctan(x)+ C1= arctan(y)+ C2" (also, do NOT use "x" and "X" interchangeably- they are different symbols).

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]

ehild
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#4
Sep1-10, 11:47 PM
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1+x^2 dy/dx = 1+y^2, differential equation! a little Help here!


Quote Quote by HallsofIvy View Post
What is true is that
[tex]tan(A+ B)= \frac{tan(A)+ tan(B)}{1+ tan(A)tan(B)}[/tex]

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
Hurkyl
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#5
Sep2-10, 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+x2, but I suppose we should let him clarify rather than assuming one way or another.
ehild
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Sep2-10, 12:36 AM
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Quote Quote by Hurkyl View Post
The joys of sloppy use of parentheses.

It looks like the OP meant to have a pair of parentheses around 1+x2, but I suppose we should let him clarify rather than assuming one way or another.
Yes, Hurkyl, but I answered to HallsofIvy, not to the OP. This custom of not using parenthesis is quite frequent on the Forum and I think we should do something against it, instead of reading minds.

ehild
HallsofIvy
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Sep2-10, 07:17 AM
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Quote Quote by ehild View Post
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]
Thanks. I keep getting that wrong!

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
Unfortunately people on this board tend to be so sloppy with parentheses I just automatically assumed that [itex](1+ x^2) dy/dx= 1+ y^2[/itex] was intended, but yes, the correct interpretation of what was written is as you say.


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