# Simple diff.eq

1. Dec 5, 2005

### sony

d/dx(y*x^2)=x^2/(1+x^2)

I guess I can't write this as:
dyx^2=x^2dx/(1+x^2)
dy=dx/(1+x^2)
beacuse I don't get the right answer...

So what do I do?

2. Dec 5, 2005

### Tide

You can't just divide out the $x^2$ in the differential. I suggest direct integration of the DE with respect to x.

3. Dec 5, 2005

### sony

well, it's the d/dx(y*x^2) that confused me.
we have only learned to solv the types:
dy/dx=f(x)g(y)

what is direct integration?

Thanks

4. Dec 5, 2005

### Tide

Direct integration means this:
$$\int_{x_0}^{x} \frac {d}{dx'}\left(x'^2 y\right) dx' = \int_{x_0}^{x} \frac {x'^2}{1+x'^2} dx'$$

5. Dec 6, 2005

### sony

Then I don't think we have learned about direct integration...

6. Dec 6, 2005

### dextercioby

Well, Tide said that, simply

$$\int du = u +\mathcal{C}$$

You must understand what i've written, else why attempt to solve diff. eqns. ?

Daniel.

7. Dec 6, 2005

### sony

okey, I understand what you just wrote :P

but I still don't know how to solve my problem...

8. Dec 6, 2005

### HallsofIvy

Staff Emeritus
Tide's point was just that you need to divide both sides by x2 to get simply $dy= \frac{dx}{1+x^2}$.

"Direct integration" just meant regular integration!

$$y= \int dy= \int \frac{dx}{1+ x^2}[/itex] Last edited: Dec 6, 2005 9. Dec 6, 2005 ### sony Oh, but I didn't think I could do that when y*x^2 was encapsulated by parantheses. When I solve what you posted I get y=tan^-1(x) but the answer is supposed to be: (pi/4)(1/x^2)-(1/x)-tan^-1(x)/x^2 (the pi/4 comes from the start value...) 10. Dec 6, 2005 ### HallsofIvy Staff Emeritus Oops! Sorry about that. I misread the original equation! It's not that the yx2 was "encapsulated by parentheses" but that here, specifically, the parentheses mean that the entire function yx2 is being differentiated. I read it as $x^2\frac{dy}{dx}$ rather than $\frac{dx^2y}{dx}$. Given that, Tide meant that [tex]x^2y= \int d(x^2y)= \int \frac{x^2dx}{x^2+ 1}$$.

That right hand side is a little more complicated. First divide it out:
$$\frac{x^2}{x^2+1}= 1- \frac{1}{x^2+1}[/itex] and it becomes easy to integrate. Last edited: Dec 6, 2005 11. Dec 6, 2005 ### sony Oh, thanks! Now I see it :) 12. Dec 6, 2005 ### sony Bah, sorry. I don't see what the integral of "dx^2y" evaluates to :( 13. Dec 6, 2005 ### HallsofIvy Staff Emeritus Fundamental Theorem of Calculus! [tex] \int d(Anything)= Anything+ C$$

$$\int d(x^2y)= x^2y + C$$

14. Dec 6, 2005

### sony

Oh! sorry! I see it! bah

hehe, the yx^2 messed up my head :P