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

[shseo0315]

Homework Statement

1+x^2 dy/dx = 1+y^2

Homework Equations

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:
$$\tan^{-1}y=\tan^{-1}x+\tan^{-1}(d)$$
Take tan of both sides and use the double angle formula for tan.

 

[HallsofIvy]

Homework Statement

1+x^2 dy/dx = 1+y^2

Homework Equations

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
$$tan(A+ B)= \frac{tan(A)+ tan(B)}{1+ tan(A)tan(B)}$$
with A= arctan(x) and B= C, tan(A)= x so
$$y= tan(arctan(x)+ C)= \frac{x+ tan(C)}{1+ tan(C)x}$$

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',
$$y= \frac{x+ C'}{1+ C'x}$$

 

[ehild]

What is true is that
$$tan(A+ B)= \frac{tan(A)+ tan(B)}{1+ tan(A)tan(B)}$$

Sorry, HallsofIvy, it is not true. The correct formula is:

$$tan(A+ B)= \frac{tan(A)+ tan(B)}{1- tan(A)tan(B)}$$

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

$$y=\frac{x}{1+cx}$$:


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]

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]

Sorry, HallsofIvy, it is not true. The correct formula is:

$$tan(A+ B)= \frac{tan(A)+ tan(B)}{1- tan(A)tan(B)}$$

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

$$y=\frac{x}{1+cx}$$:


ehild

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