Solve the following D.E., \[\frac{dy}{1+y^2}=\frac{dx}{1+x^2}\], obtaining the result in algebraic form

- anonymous

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- katieb

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- anonymous

the solution given by the text is \[y=x+C(1+xy)\]

- lgbasallote

i believe you should cross multiply first \[(1 + x^2)dy - (1+y^2)dx = 0\]

- myininaya

You just integrate both sides of that equation you have @ebbflo

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## More answers

- myininaya

You already have the variables separated

- UnkleRhaukus

\[\int \frac{1}{1+z^2}\text dz=\arctan z+c\]

- myininaya

Is your trouble writing it as an algebraic expression?

- anonymous

I see all of your points and have tried those methods but do not obtain the given answer....

- myininaya

If so.. Don't forget you only need one constant (and just put it on one side of your equation) Remember the inverse of tan inverse which is tan lol also remember the formula tan(a+b)=(tan(a)+tan(b))/(1-tan(a)*tan(b)) Then remember that tan(constant) is still a constant)

- myininaya

Remember @ebbflo it does say write as an algebraic expression.

- myininaya

You get trigonometric expression right after integration

- myininaya

You want to transform that to an algebraic one using the hint I just gave you

- myininaya

Also recall tan(arctan(p))=p

- myininaya

You will use that too

- anonymous

thanks, I understand all the points you have made

- myininaya

What about the solution? Have you got the desired solution?

- myininaya

Or do you need more help?

- anonymous

no, got it, i had the solution, just don't quite understand "why" the book's solution is written in the form it is... the first copyright is 1943, maybe its a style thing...

- anonymous

@myininaya , you can use Latex so that your math text is more clear

- myininaya

That one formula is a trig formula The expansion for tan(a+b) let me know if you need anything else

- anonymous

as i said, I had an equivalent solution, I was just curious as to why the book chose to write the solution in the form it did...

- myininaya

Was your form algebraic?

- myininaya

Can you write what you have?

- anonymous

I have the solution given, I think the book just wanted a form with one arbitrary constant, i generally don't like my solutions to have "y" on the LHS and RHS when it is not absolutely necessary...it was really ore of a "style" question, sorry I should have specified

- myininaya

If the equation can be written in the form the book gives, then your answer is right I got the same answer your book got I wrote it in a different form but it is still correct

- anonymous

yes I understand, thank you...as I said I was questioning the style, and now that I think of it it was probably that the books form only take a single line of text whereas mine did not...;)

- UnkleRhaukus

differential equations will often have \(y\) on both sides of the solution, but that is alright. we were only trying to get rid of derivatives replace them with constants

- anonymous

@unkle agreed, but it is my preference not to do so when not necessary

- anonymous

I should have been more clear, I was not so much looking for "help" with the problem, but a different perspective on how another might write the solution

- UnkleRhaukus

these are some solution to differential equations ,

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- anonymous

in order to make a decision as to what a student may find more clear

- UnkleRhaukus

your solution must include as many arbitrary constants as the order of the differential equation

- anonymous

exactly uncle, in those solution you provided, the "y" cannot be written explicitly in terms of "x", at least not in a single expression

- UnkleRhaukus

well, it can but i just looks awful , these are 'neater'

- UnkleRhaukus

looking at (e)

- anonymous

yeah that one can but what about (c)?

- anonymous

(e) more or less is already, not much more to do on that one...

- UnkleRhaukus

i think it is clearer to write (c) like \[x = y^2(\ln y + c)\] i dont know why they chose the form they did

- UnkleRhaukus

implicit solutions are fine for DE's

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