Differentiating y*x^2: x^2/(1+x^2)

[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?

 

[Tide]

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

 

[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

 

[Tide]

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

 

[sony]

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

 

[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.

 

[sony]

okey, I understand what you just wrote :P

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

 

[HallsofIvy]

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?

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!

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

 

[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...)

 

[HallsofIvy]

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...)

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
$$x^2y= \int d(x^2y)= \int \frac{x^2dx}{x^2+ 1}$$.

That right hand side is a little more complicated. First divide it out:
[tex]\frac{x^2}{x^2+1}= 1- \frac{1}{x^2+1}[/itex]<br /> and it becomes easy to integrate.[/tex]

 

[sony]

Oh, thanks! Now I see it :)

 

[sony]

Bah, sorry. I don't see what the integral of "dx^2y" evaluates to :(

 

[HallsofIvy]

Bah, sorry. I don't see what the integral of "dx^2y" evaluates to :(

Fundamental Theorem of Calculus!

$$\int d(Anything)= Anything+ C$$

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

 

[sony]

Oh! sorry! I see it! bah

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