What is the solution to the differential equation dy/dx=(y-y2)/x for x≠0?

In summary, the conversation discusses solving a differential equation and the use of integration by parts and partial fractions. It also includes a question about the derivative and integral of ln(x). The correct derivatives and integrals are provided and explained.
  • #1
mmekosh
3
0

Homework Statement



Solve the differential equation: dy/dx=(y-y2)/x , for all x[tex]\neq[/tex] 0

Homework Equations



Integration by Parts: [tex]\int[/tex] u dv = u v - [tex]\int[/tex] v du

[tex]\int[/tex]lnx= 1/x + C
[tex]\int[/tex] (1/x) = lnx + C
dy/dx lnx = 1/x
dy/dx 1/x = lnx

The Attempt at a Solution



dy/(y-y2)=dx/x

[tex]\int[/tex] 1/(y-y2) dy = [tex]\int[/tex] 1/x dx

[tex]\int[/tex] (1/y)(1/(1-y))dy = lnx + C

(Integration by parts)
u=1/x dv=(1/(1-y))dy
du=lnydy v= -ln(1-y)dy

-lny / y + [tex]\int[/tex] lny ln(1-y) dy

And then if I continue and do integration by parts again, it just goes back to the original integral.
 
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  • #2
Instead of trying to do this by integration by parts, use the technique of partial fractions. In a nutshell you want to rewrite your integral on the left side:
[tex]\int \frac{dy}{y(1 - y)}[/tex]

with the fraction broken up into two separate fractions, like so:
[tex]\int \left[\frac{A}{y} + \frac{B}{1 - y}\right]dy[/tex]

What you need to do is to find constants A and B so that 1/(y(1 - y)) is identically equal to A/y + B/(1 - y).
 
  • #3
mmekosh said:
[tex]\int[/tex]lnx= 1/x + C
[tex]\int[/tex] (1/x) = lnx + C
dy/dx lnx = 1/x
dy/dx 1/x = lnx
Hello!
I don't mean to butt in, but can someone verify that:
[tex]\frac{d}{dx}(\frac{1}{x}) = \ln{x}?[/tex]
Also, the integral:
[tex]\int{\ln{x}} \left \left dx = \frac{1}{x}?[/tex]
I'm confused...
Many thanks.
 
  • #4
nobahar said:
Hello!
I don't mean to butt in, but can someone verify that:
[tex]\frac{d}{dx}(\frac{1}{x}) = \ln{x}?[/tex]
Also, the integral:
[tex]\int{\ln{x}} \left \left dx = \frac{1}{x}?[/tex]
I'm confused...
Many thanks.

I think you've confused the concept of integral with that of derivative as

[tex]\frac{d}{dx}(\ln(x)) = \frac{1}{x}[/tex]
[tex]\int{\frac{1}{x}} \left \left dx = \ln(x)+C[/tex].

And yeah both are correct.

AB
 
  • #5
nobahar said:
Hello!
I don't mean to butt in, but can someone verify that:
[tex]\frac{d}{dx}(\frac{1}{x}) = \ln{x}?[/tex]
No, no one can verify that- it's not true. What is true is that
[tex]\frac{d}{dx}(\frac{1}{x})= -\frac{1}{x^2}[/tex]
because [itex]\frac{1}{x}= x^{-1}[/itex] and the derivative of [itex]x^n[/itex] is [itex]nx^{n-1}[/itex].

Also, the integral:
[tex]\int{\ln{x}} \left \left dx = \frac{1}{x}?[/tex]
I'm confused...
Many thanks.
No, that's also not true. [itex]\int ln(x) dx= ln(x)(x- 1)+ C[/itex]

You have these both backwards: [itex]d(ln(x))/dx= 1/x[/itex] and [itex]\int 1/x dx= ln(x)+ C[/itex].
 
  • #6
Thanks both Altabeh and Halls.
I didn't consider them to be true, hence why I said I was confused. It was from the OP's relevant equations section.
 

1. What is the definition of a derivative of an integral?

The derivative of an integral is a mathematical concept that represents the rate of change of the integral with respect to its variable. It is calculated by taking the limit of the difference quotient of the integral as the interval approaches zero.

2. How is the derivative of an integral calculated?

The derivative of an integral can be calculated using the fundamental theorem of calculus, which states that the derivative of an integral is simply the original function evaluated at the upper limit of integration.

3. What is the relationship between the derivative and the integral?

The derivative and the integral are inverse operations of each other. This means that the derivative of an integral is the original function and the integral of a derivative is the original function, as long as the constant of integration is taken into account.

4. Can the derivative of an integral be negative?

Yes, the derivative of an integral can be negative. This means that the rate of change of the integral is decreasing with respect to its variable. For example, the derivative of the integral of a decreasing function will be negative.

5. How are derivatives of integrals used in real-world applications?

Derivatives of integrals are used in various fields of science and engineering to model and analyze real-world phenomena. For example, in physics, they are used to calculate the velocity and acceleration of objects, and in economics, they are used to model supply and demand curves. They are also used in optimization problems to find the maximum or minimum of a function.

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