Question

Solve:\[\frac{dy}{dx} =\frac{1+y^2}{1+x^2}\]

Answer 1

1. Take (1+y^2) in LHS and dx in RHS

Answer 2

It's seperable

Answer 3

So, is this a separable eq?

Answer 4

Yep.

Answer 5

We get:

\[\large{\cfrac{dy}{1+y^2} = \cfrac{dx}{1+x^2}}\]

Now integrate both sides

Answer 6

Can you find a way to get only x's on one side and only y's on the other?

Answer 7

\[\checkmark \checkmark\]

Answer 8

Remember this:

\[\large{\int \cfrac{dy}{1+y^2} = \tan^{-1}y + C}\]

Answer 9

Now can you solve this after this ?

Answer 10

so, then, ya just integrate wrt the variables on each corresponding side, you got this

Answer 11

I cant really help you I dont know this yet Im sorry

Answer 12

I posted this up for fun :P

Answer 13

It's symmetric

Answer 14

This method is called variable separation :D

Answer 15

Well done, @vishweshshrimali5

Answer 16

oh, I thought this was a bit basic for you

Answer 17

It's the RHS i'm confused with.

Answer 18

what exactly?

Answer 19

Why?

Answer 20

\[tan^{-1}(y) = \tan^{-1}(x) + C\] but it's not \(C\) it's written as \(\tan^{-1}(C)\)

Answer 21

See you can write it as you wish

Answer 22

That is incorrect notation

Answer 23

Constant will remain constant. lol

Answer 24

thank you for trying to entertain us but I know another way you can entertain me

Answer 25

it has to be +C

Answer 26

C is basically just a constant.

\[\large{\tan^{-1}(C)}\] would also be a constant

Answer 27

and that is by teaching me how to do this
;)

Answer 28

You can modify the integration constant as the need arises to make the final equation more presentable :)

You can write: C or C^2 or C^3 or sqrt(C) or C/2 and so on

Answer 29

But that is not correct, you would have to say \(tan^{-1}(x)+tan^{-1}(C)\) you cannot just say of C alone

Answer 30

C can be anything you want, but you must include it in someway

Answer 31

The solution here is written as \[\tan^{-1}(y) = \tan^{-1}(x) +\tan^{-1}(C)\]\[y=\tan(\tan^{-1}(x)+\tan^{-1}(C)) = \frac{x+C}{1-Cx}\]

Answer 32

@Jhannybean You can replace C by some other constant say C'

And I have chosen:
C' = tan^(-1) C

Answer 33

Ohh, ok makes sense, makes sense.

Answer 34

I agree with the first step, but not sure how they did the simplification to that fraction, there must be a trig sub I am unaware of

Answer 35

It does not matter whether I write C or C' as both are some constants whose value is unknown to me.

Answer 36

I was just curious if there was another method in solving this.

Answer 37

As for the simplification:

It is just a simple inverse trigonometry formula

Answer 38

@Kainui

Answer 39

could always use variation of parameters or similar, but the easiest way is the best way here

Answer 40

You can always use substitution

Answer 41

Substitute:

\[\large{x = \tan (u)}\]
\[\large{\implies u = \tan^{-1} x}\]

\[\large{\implies du= \cfrac{1}{1+x^2} dx}\]

Similarly:

\[\large{y = \tan (v)}\]

\[\large{\implies dv = \cfrac{1}{1+y^2}dy}\]

Answer 42

Now your integrals would simplify to:

\[\large{\int du = \int dv}\]

\[\large{\implies u = v + c}\]

Replace them with \(\tan^{-1} x\) and \(\tan^{-1} y\) and you would get the same answer

Again you would have to replace c by \(\tan^{-1} c\)

Answer 43

Though, just an important point:

I just use the same basic concept in a different way. This method is called substitution method.

Answer 44

and why exactly would we have to replace c?

Answer 45

Its just to make your equation more presentable

Answer 46

to make it look pretty usually,

Answer 47

So it's not a crime to leave it as \(C\).... it would just be considered ambiguous?

Answer 48

nope, C is perfectly acceptable

Answer 49

Let me ask you something:

Which one do YOU like more ?

\[\cfrac{1}{3} x + \cfrac{1}{2} y =1\]

or

\[2x + 3y = 6\]

Similarly:

\[\large{x^2 + y^2 = 1}\] or \[\large{r = 1}\]

Answer 50

I had understood the rest of the integration process and how to solve this, I was just simply stuck on the constant part, heh.

Answer 51

Oh I see @FibonacciChick666 :)

Answer 52

Making it pretty and presentable is all there is, now you can use any arbitrary constant but if you do not put it, It WILL be incorrect, so just remember that

Answer 53

Use trig addition formula for \(\tan\).

Answer 54

Yeah ofc.

Answer 55

I never heard of a trig addition formula???(or if I did, I totally forgot)

Answer 56

Sum angle formula

Answer 57

oh yeah, that makes sense.\[\tan(\alpha) +\tan(\beta) = \frac{\tan(\alpha) +\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}\]

Answer 58

\(\tan^{-1}(C_1)\) is as much an arbitrary constant as \(C\)

Answer 59

@vishweshshrimali5 It's been too long since I've seen you! Good to see you're still on top of integrals!

@Jhannybean Another example where choosing +C as something else as an example of its utility is in this common integral: \[\Large y= \int\limits \frac{1}{x}dx \\ \Large y= \ln x + C\] if we choose C instead to be ln(C) we have: \[\Large y = \ln x + \ln C = \ln(xC) \\ \Large e^y=Cx\] Which you might be more comfortable rearranging around with completely different variables as \[\Large y=Ce^x\]

Answer 60

Hi @Kainui !

College really makes things difficult in beginning ;) But, now I'm back :D

Answer 61

\(\color{blue}{\text{Originally Posted by}}\) @ganeshie8
\(\tan^{-1}(C_1)\) is as much an arbitrary constant as \(C\)
\(\color{blue}{\text{End of Quote}}\)
Technically it is restricted to a certain image and may have multiple solutions.

But where di \(\tan^{-1}(C_1)\) come frome anyway?

Answer 62

That's what I was wondering in the first place.

Answer 63

@wio We just replaced an arbitrary constant with another arbitrary constant

Answer 64

Though tan^-1(C) isn't arbitrary technically...

Answer 65

∫1x2dx=2∫1y(y2−3)dy

Answer 66

it is restricted

Answer 67

-1/x+c=ln(y^2-3)
-1/1+c=ln(2^2-3)
-1+c=ln(1)
-1+c=0

Answer 68

recall the domain of arctan(x) fib

Answer 69

what restrictions are you guys talking about ? :/

Answer 70

domain is, but what we end with isn't then

Answer 71

@FibonacciChick666 arctan(x) is defined for all real values of x

Answer 72

em that made no sense, let me say, ok if we just had C we could make it equal to 8 right? but tan^-1{C} can never equal 8

Answer 73

we are eliminating answers by using tan^-1 in my opinion

Answer 74

tan^(-1) C can never be 8...uhm. why ?

Answer 75

because it is only defined on y=-pi/2, pi/2

Answer 76

got you point, so are you saying the restriction is going away because we're removing the arctan before reaching final solution ?

Answer 77

@FibonacciChick666 You are right, but if we do that then we would be getting into complex analysis for the full answer, but it's nice to have a simple, real answer to evaluate as well.

Answer 78

What @Kainui said ^^^ :D

Answer 79

I am saying that since arctan has a restricted range, we are eliminating, so it would be better and all inclusive just to put C or D or whichever variable you want instead

Answer 80

http://www.cliffsnotes.com/assets/11440.jpg

Answer 81

The "function" arctan has the range (-pi/2, pi/2) but that does NOT mean that tan(pi) won't exist...

Answer 82

If you make an unnecessary restriction on \(C\), then you are eliminate potential solutions.
If you make get rid of necessary restrictions for \(C\), then you allow false solutions.

Answer 83

^what he said, since the restriction of the range of inv. tan isn't needed, it removes solutions

Answer 84

hmm...

Answer 85

\[
\tan^{-1}(y)=\tan^{-1}(x)+C\\
y = \tan(\tan^{-1}(x)+C) = \frac{x+\tan(C)}{1-x\tan(C)} = \frac{x+C'}{1-C'x}
\]I don't think this is necessarily a bad answer.

Answer 86

We can write \[\Large y = \tan ( \arctan(x) +C)\] However in a physical situation y taking on complex numbers is meaningless.

Answer 87

I mean, personally, I would just leave it there^ No need to simplify it. It's done enough

Answer 88

and so long as it is +C and not +tan^-1C, I will agree

Answer 89

Yeah, besides indefinite integrals don't really mean anything. In real life you would be solving a problem with initial conditions and such or simply be able to set it to whatever you like just like we arbitrarily assign where 0 potential energy is. It can be at infinity, or ground level, it doesn't matter.

Answer 90

but this is math. We never have REAL WORLD problems. NEVER! That's just absurd. **

Answer 91

thanks for clearing that up :)

Answer 92

Yeah that's my point, since it's not the real world, let's cut out this ugly answer you wrote and rewrite it with restrictions so it looks all pretty and mathematically elegant. =P

Answer 93

lol thats so true :P
Okay, so aren't we committing the same mistake with C and lnC business or it doesn't matter here because ln(C) produces all real numbers ?

Answer 94

If you were to just ignore the existence of indefinite integrals all together...\[
\int_C^x \frac{dt}{1+t^2} = \tan^{-1}(x) - \tan^{-1}(C)
\]

Answer 95

Yeah, ln(C) is just as good as plain C because they have the same range. For every value that C takes on we have an exact 1 to 1 for every value ln(C) can take on. It just seems weird because the domain of ln(C) is different.

Answer 96

agreed