Hmm...I'm familiar with this type of substitution but never came to know its called Weierstrauss Substitution.
For some reason of the many calculus texts I gazed upon there is only one that mentions its name or even the techinque.

2. Originally Posted by ThePerfectHacker
For some reason of the many calculus texts I gazed upon there is only one that mentions its name or even the techinque.
Heard of it, never knew what it was.

-Dan

3. Didn't anyone take a look at this one. Solve the differential equation
$
(xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy=0
$

Didn't anyone take a look at this one. Solve the differential equation
$
(xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy=0
$
Yes, I did. But I am not very talented when it comes to differencial equations. Thus far I did not find any way to do this. Sadly, no integrating factor exists as mentioned in the other post.

5. Originally Posted by ThePerfectHacker
Yes, I did. But I am not very talented when it comes to differencial equations. Thus far I did not find any way to do this. Sadly, no integrating factor exists as mentioned in the other post.
OK, the section of the book in which the question lies has the other questions done by substitution but I couldn't do this one.

Taking $x=r\,\cos\theta,\,y=r\,\sin\theta$ where $\theta$ & $r$ are variables
So, $x^{2}+y^{2}=r^{2}$
and $\tan\theta\,=\,\frac{y}{x}$
On differentiating the above two equations we can substitute $xdy+ydx$ & $xdy-ydx$ in the differential equation.

So does this help?

Didn't anyone take a look at this one. Solve the differential equation
$
(xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy=0
$
First, is this equation correct? Usually if there's an "e" in there it's $e^x$ or something.

Let's try:
$(xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy=0$

$x^{2}y \frac{dy}{dx}-xy^{2}=-\frac{e}{x^{3}}$

$\frac{dy}{dx}-\frac{y}{x}=-\frac{e}{x^{4}}y^{-1}$ or the form $\frac{dy}{dx} + P(x)y = Q(x)y^{n}$

This kind of non-linear first order differential equation is a "Bernoulli" equation with n = -1.

Let $v = y^{2}$. Then $\frac{dv}{dx} = 2y \frac{dy}{dx}$, or $\frac{dy}{dx} = \frac{1}{2\sqrt{v}} \frac{dv}{dx}$.

Thus:
$\frac{1}{2\sqrt{v}} \frac{dv}{dx} - \frac{\sqrt{v}}{x} = -\frac{e}{x^4}\frac{1}{\sqrt{v}}$

Thus
$\frac{dv}{dx} - \frac{2}{x}v = -\frac{2e}{x^4}$

which is now a linear first order differential equation, which you can solve using any method you wish. (Remember to back-substitute to get y(x)!)

-Dan

I've tried the polar coordinate substitution you suggested, but I'm not getting any "joy" out of it.

7. Originally Posted by topsquark
First, is this equation correct? Usually if there's an "e" in there it's $e^x$ or something.
I'm sure that is what's written in the book.

Originally Posted by topsquark
Let's try:
$(xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy=0$

$x^{2}y \frac{dy}{dx}-xy^{2}=-\frac{e}{x^{3}}$

$\frac{dy}{dx}-\frac{y}{x}=-\frac{e}{x^{4}}y^{-1}$ or the form $\frac{dy}{dx} + P(x)y = Q(x)y^{n}$

This kind of non-linear first order differential equation is a "Bernoulli" equation with n = -1.

Let $v = y^{2}$. Then $\frac{dv}{dx} = 2y \frac{dy}{dx}$, or $\frac{dy}{dx} = \frac{1}{2\sqrt{v}} \frac{dv}{dx}$.

Thus:
$\frac{1}{2\sqrt{v}} \frac{dv}{dx} - \frac{\sqrt{v}}{x} = -\frac{e}{x^4}\frac{1}{\sqrt{v}}$

Thus
$\frac{dv}{dx} - \frac{2}{x}v = -\frac{2e}{x^4}$

which is now a linear first order differential equation, which you can solve using any method you wish. (Remember to back-substitute to get y(x)!)
Thanks for the nice explanation.

Originally Posted by topsquark
I've tried the polar coordinate substitution you suggested, but I'm not getting any "joy" out of it.
Do you think I do?

I'm sure that is what's written in the book.

Thanks for the nice explanation.

Do you think I do?
You're welcome. Actually the solution surprized me. I didn't know until now that Bernoulli equations can be solved for negative n. So thank you for the opportunity!

-Dan

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