Separation of variables question

[JamesGoh]

Homework Statement

Solve the following separable equation

[itex]\frac{dy}{dx} = \frac{y}{x(x-1)}[/itex]

Homework Equations

[itex]\int\frac{1}{x}dx=ln(x)[/itex]

The Attempt at a Solution

See attachment

Im getting [itex]y=(\frac{x-1}{x})^{c}[/itex] as my answer when in fact the answer the tutor gave is [itex]y=c(\frac{x-1}{x})[/itex]

not sure what I am missing

 

[Tomer]

I don't understand why you say that:
[itex]\int\frac{dx}{x(1-x)} = \int \frac{cdx}{1-x}- \int \frac{cdx}{x}[/itex]

What is this c?
You should be writing:
[itex]\int\frac{dx}{x(1-x)} = \int (\frac{a}{1-x} + \frac{b}{x})dx[/itex]
and find the appropriate a and b.

 

[JamesGoh]

I don't understand why you say that:
[itex]\int\frac{dx}{x(1-x)} = \int \frac{cdx}{1-x}- \int \frac{cdx}{x}[/itex]

What is this c?
You should be writing:
[itex]\int\frac{dx}{x(1-x)} = \int (\frac{a}{1-x} + \frac{b}{x})dx[/itex]
and find the appropriate a and b.

c is a constant value

a and b would be the same value, since the numerator on the [itex]\int\frac{dx}{x(1-x)} [/itex] is 1

you just have to treat [itex] (\frac{a}{1-x} + \frac{b}{x}) [/itex] like adding fractions and you'll realize why I simply put c on the top of both fractions

 

[Tomer]

So if you know c = 1, why do you keep on dragging it?
It's not the "c" in the final answer.
The "c" in the final answer comes as a result of the integration constant.

[itex]\int \frac{dx}{x(x-1)} = ln(\frac{x}{x-1}) + C[/itex]

EDIT: I just reread your post - maybe you weren't aware that your c=1? Of course I realize that a and b have the same values (actually opposite values) - but you should also notice that a = -b = 1.

 

[lurflurf]

You added a false constant and omitted the actual constant.

dy/y=dx/[x(x-1)]
look like log write as log
d log(y)=d log(1-1/x)
integrate
y=C (1-1/x)

1/[x(x-1)]=1/(x-1)-1/x=(1/x)[1/(1-1/x)-(1-1/x)/(1-1/x)]
=(1/x^2)/(1-1/x)=(1-1/x)'/(1-1/x)=d log(1-1/x)
or
1/[x(x-1)]=1/(x-1)-1/x=d log(x-1)-d log(x)=d log(1-1/x)

 

[JamesGoh]

So if you know c = 1, why do you keep on dragging it?
It's not the "c" in the final answer.
The "c" in the final answer comes as a result of the integration constant.

[itex]\int \frac{dx}{x(x-1)} = ln(\frac{x}{x-1}) + C[/itex]

EDIT: I just reread your post - maybe you weren't aware that your c=1? Of course I realize that a and b have the same values (actually opposite values) - but you should also notice that a = -b = 1.

I was aware that c=1 from adding the fractions together

I explored the idea of using integration constant, however I backed out because I wouldn't be able to find an inverse natural log of it.

Im guessing cause C (integration constant ) is an arbitrary number, we can simply rewrite it as ln(C) ?

 

[Tomer]

I was aware that c=1 from adding the fractions together

I explored the idea of using integration constant, however I backed out because I wouldn't be able to find an inverse natural log of it.

Im guessing cause C (integration constant ) is an arbitrary number, we can simply rewrite it as ln(C) ?

You could do that, or you could look at it like this:

[itex]ln(y) = ln(\frac{x}{x-1}) + c => y(x) = e^{ln(\frac{x}{x-1}) + c} =>

y(x) = e^{c}ln(\frac{x}{x-1}) => y(x) = Cln(\frac{x-1}{x})^{-1} =>
y(x) = -Cln(\frac{x-1}{x}) =>
y(x) = Cln(\frac{x-1}{x})[/itex]

The example shows how C can alter it's value (c --> e^c --> -e^c). However I didn't bother to "rename" it every time, because it's clear that it's just a constant.

You'll always have C popping in a first order PDE, and that's because you'll always have a constant of integration. That's always the source of c.