First order differential equation

[dud6913]

Homework Statement

Hi,

If x^2+1=y/(x-y'), where y'=dy/dx, find dy/dx

I have tried so many ways, but I cannot seem to get the correct answer.

The answers I have got previously are:
i)
x² + 1 = y/(x - y')
(x² + 1)(x - y') = y
x(x² + 1) - y'(x² + 1) = y
x(x² + 1) - y = y'(x² + 1)
dy/dx = x - y/(x² + 1)

ii)

Switch divisors:
x – dy/dx = y / (x² + 1)

Now switch sides:
dy/dx = x - y / (x² + 1)
dy/dx= (x³ + x – y) / (x² + 1)

However, both of them do not seem correct.

Please help!

Any suggestions are welcome.

Also, I think that this is a linear ode. Even if i was to think of this as separable function, I would not have a clue how to get y' on the other side, as I would have to expand the left side with (x-y').

Ta

Homework Equations

The Attempt at a Solution

 

[hunt_mat]

How about saying that:
<br /> x-\frac{dy}{dx}=\frac{y}{1+x^{2}}<br />
Then from there it's an easy matter to obtain dy/dx

 

[dud6913]

Hi,

Yeah i tried that before. If you can see under i), I have done that and got dy/dx = x - y/(x² + 1). However, that is not the correct answer.

 

[hunt_mat]

Then I would suggest that the answer in the book is wrong. What you wrote in 1) is correct.

 

[dud6913]

I think that I am supposed to calculate dy/dx by thinking that the equation given, is separable equation or linear. This is a first order differential equation, thus the answer cannot be this simple...

Thanks for your contribution though.

 

[hunt_mat]

In that case I don't know what your lecturer wants. Possibly to solve the equation in terms of x and then compute dy/dx to obtain it as a function of x. If this is the case then the equation may be solved via the integrating factor method.
<br /> \frac{dy}{dx}+\frac{y}{1+x^{2}}=x<br />
where the integrating factor is:
<br /> e^{\tan^{-1}x}<br />

 

[HallsofIvy]

You say the answers you got were
\frac{dy}{dx}= x- \frac{y}{x^2+ 1}
and
\frac{dy}{dx}= \frac{x^3+ x- y}{x^2+1}

Those are, of course, exactly the same:
x- \frac{y}{x^2+1}= \frac{x(x^2+1)}{x^2+1}- \frac{y}{x^2+ 1}
= \frac{x^3+ x}{x^2+ 1}- \frac{y}{x^2+ 1}= \frac{x^3+ x- y}{x^2+ 1}

But they are "answers" to what question?

As Hunt Mat says, the equation is linear:
\frac{dy}{dx}= x- \frac{y}{x^2+ 1}
can be written as
\frac{dy}{dx}+\frac{1}{x^2+ 1}y= x
and an integrating factor can be found as he says.

 

[dud6913]

All that I know, is that I have to find dy/dx from the equation x^2+1=y/(x-dy/dx).

The multiple choice answers are the following:

a)dy/dx= 2(x+y)^2+ (x/y)
b)dy/dx=(2x-y)^2+(x/y)
c)dy/dx=2(x-y)^2+(y/x)
d)dy/dx=2(x-y)^2+(y/x)
e)none of the above

I have tried the advice hunt_mat has given me, but the answer I acquired following that process has no resemblance to any of these answers.

I know that I have not supplied enough information, but that is all I have been given.

Hope this clarifies it for you guys, so you can assist me...

I have spent a lot of time doing this question, but maybe the answer is e) :D

Thank you

 

[dud6913]

Also, I am aware of the fact that the equation is a separable 1st order ordinary differential equation, but by substituting the factor e^tan-1(x), the answer will be in "e" format, which has no relevance to the given answers. There has to be a trick, or maybe I am just confusing myself even more.

 

[hunt_mat]

Whay about none of the above?

 

[dud6913]

I just don't think that the answer would be that obvious... Maybe I am wrong.

Thanks for the input though guys, much appreciated.