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

shseo0315

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!

hunt_mat

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

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

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

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

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

Thanks. I keep getting that wrong!

By the way, the original equation

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.