First order differential equation

In summary, the student is struggling to solve an equation that is stated to be linear. They have tried different methods but have not been able to find the correct answer. They think that the answer is e) but they are not sure. They appreciate any input that can help them.
  • #1
dud6913
11
0

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

 
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  • #2
How about saying that:
[tex]
x-\frac{dy}{dx}=\frac{y}{1+x^{2}}
[/tex]
Then from there it's an easy matter to obtain dy/dx
 
  • #3
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.
 
  • #4
Then I would suggest that the answer in the book is wrong. What you wrote in 1) is correct.
 
  • #5
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.
 
  • #6
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.
[tex]
\frac{dy}{dx}+\frac{y}{1+x^{2}}=x
[/tex]
where the integrating factor is:
[tex]
e^{\tan^{-1}x}
[/tex]
 
  • #7
You say the answers you got were
[tex]\frac{dy}{dx}= x- \frac{y}{x^2+ 1}[/tex]
and
[tex]\frac{dy}{dx}= \frac{x^3+ x- y}{x^2+1}[/tex]

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

But they are "answers" to what question?

As Hunt Mat says, the equation is linear:
[tex]\frac{dy}{dx}= x- \frac{y}{x^2+ 1}[/tex]
can be written as
[tex]\frac{dy}{dx}+\frac{1}{x^2+ 1}y= x[/tex]
and an integrating factor can be found as he says.
 
  • #8
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
 
  • #9
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.
 
  • #10
Whay about none of the above?
 
  • #11
I just don't think that the answer would be that obvious... Maybe I am wrong.

Thanks for the input though guys, much appreciated.
 

1. What is a first order differential equation?

A first order differential equation is a mathematical equation that relates an unknown function to its derivative. It can be expressed in the form of dy/dx = f(x) where y is the unknown function and f(x) is some known function.

2. What is the difference between an ordinary differential equation and a partial differential equation?

An ordinary differential equation involves only one independent variable, while a partial differential equation involves multiple independent variables. This means that the unknown function in a partial differential equation can have multiple partial derivatives with respect to the different independent variables.

3. How do you solve a first order differential equation?

There are several methods for solving first order differential equations, including separation of variables, integrating factors, and exact equations. The method used depends on the specific form of the equation.

4. What are the applications of first order differential equations?

First order differential equations have many applications in various fields of science and engineering, including physics, chemistry, biology, and economics. They are used to model and understand dynamic systems and processes.

5. Can all first order differential equations be solved analytically?

No, not all first order differential equations have analytical solutions. Some equations may require numerical or approximate methods for finding solutions. However, there are certain types of first order differential equations that have well-defined analytical solutions.

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