Solve Diff. Eq. by Separation of Variables

[KillerZ]

Homework Statement

Solve the given differential equation by separation of variables.

Homework Equations

[tex]\frac{dy}{dx} = \frac{xy + 2y - x - 2}{xy - 3y + x - 3}[/tex]

The Attempt at a Solution

[tex]\frac{dy}{dx} = \frac{xy + 2y - x - 2}{xy - 3y + x - 3}[/tex]

[tex] = \frac{(x + 2)(y - 1)}{(x - 3)(y + 1)}[/tex]

[tex](x - 3)(y + 1)\frac{dy}{dx} = (x + 2)(y - 1)[/tex]

[tex]\frac{(x - 3)(y + 1)}{(y - 1}\frac{dy}{dx} = (x + 2)[/tex]

[tex]\frac{(y + 1)}{(y - 1)}\frac{dy}{dx} = \frac{(x + 2)}{(x - 3)}[/tex]

[tex]\frac{(y + 1)}{(y - 1)}dy = \frac{(x + 2)}{(x - 3)}dx[/tex]

This is where I am having problems I am not sure what way to Integrate this:

[tex]\int\frac{(y + 1)}{(y - 1)}dy = \int\frac{(x + 2)}{(x - 3)}dx[/tex]

I though maybe by partial fractions but the degree of the numerator is not less than the degree of the denominator.

 

[Mark44]

No, not partial fractions. Just divide (using long division) the denominator into the numerator on each side. On the left side you'll get 1 + 2/(y - 1).

By the way, you could have skipped several steps in your work.
[tex] \frac{dy}{dx}~=~ \frac{(x + 2)(y - 1)}{(x - 3)(y + 1)}[/tex]
[tex] \frac{y + 1}{y - 1}~dy~=~ \frac{(x + 2)}{(x - 3)}~dx[/tex]
You can get to the second equation above by multiplying both sides by (y + 1)/(y - 1)dx

 

[KillerZ]

Ok I got that but now I am stuck again this time with simplifying.

Long division =

[tex]\int1 + \frac{2}{y - 1}dy = \int1 + \frac{5}{x - 3}dx[/tex]

[tex]\int1dy + \int\frac{2}{y - 1}dy = \int1dx + \int\frac{5}{x - 3}dx[/tex]

[tex]\int1dy + 2\int\frac{1}{y - 1}dy = \int1dx + 5\int\frac{1}{x - 3}dx[/tex]

I am stuck here:

[tex]y + 2ln|y-1| = x + 5ln|x-3| + c[/tex]

 

[Galadirith]

Hi KillerZ,

Polynomial long division is certainly the way I should be done, and if you ever encounter quotients like that were its a quadratic divided by a quadratic then again think Polynomial long division. However there is a way do deal with this sort of problem when it is a linear expression divided by a linear expresion, consider this example:

[tex]
\int \frac{x+5}{x-2} dx
[/tex]

now what if we wrote 5 as 7-2

[tex]
\int \frac{x+5}{x-2} dx = \int \frac{x+7-2}{x-2} dx = \int \frac{(x-2) + 7}{x-2} dx = \int 1 + \frac{7}{x-2} dx
[/tex]

and then you have a form you can integrate, this also work if there is a coefficient other than 1 for either of the x terms, but then its probably better to do the long division so as not to make a mistake, this just often is quicker to do than long division for simple cases, the irony is that this was probably discovered (if you can call it a discovery) when someone did a PLD of a quotient of this form. This I think is a lovely little "trick" that can actually be used in other places also, not just quotients :D

 

[Mark44]

Ok I got that but now I am stuck again this time with simplifying.

Long division =

[tex]\int1 + \frac{2}{y - 1}dy = \int1 + \frac{5}{x - 3}dx[/tex]

[tex]\int1dy + \int\frac{2}{y - 1}dy = \int1dx + \int\frac{5}{x - 3}dx[/tex]

[tex]\int1dy + 2\int\frac{1}{y - 1}dy = \int1dx + 5\int\frac{1}{x - 3}dx[/tex]

I am stuck here:

[tex]y + 2ln|y-1| = x + 5ln|x-3| + c[/tex]

There's not a whole lot you can do in simplification other than this:
[tex]y + ln|y-1|^2 = x + ln|x-3|^5 + c[/tex]

You're not going to be able to solve this equation for y as an explicit function of x. If you want to check your work (and you should), just differentiate implicitly and you should get back to your original differential equation.

 

[KillerZ]

Ok thanks

 

[KillerZ]

I thought I would post this here as its just another integral:

[tex]\int\frac{1}{y^{2}}dy = \int\frac{1}{e^{x}+e^{-x}}dx[/tex]

[tex]-\frac{1}{y} = \int\frac{e^{x}}{e^{2x}+1}dx[/tex]

[tex]u = e^{x}[/tex]

[tex]du = e^{x}dx[/tex]

I am not sure what to do here:

[tex]-\frac{1}{y} = \int\frac{du}{u^{2}+1}[/tex]

 

[Bohrok]

I am not sure what to do here:

[tex]-\frac{1}{y} = \int\frac{du}{u^{2}+1}[/tex]

That integral is tan^-1u + C. Make sure you learn some integrals with inverse trig functions.