- #1

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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?

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- Thread starter sony
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- #1

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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

Tide

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- #3

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we have only learned to solv the types:

dy/dx=f(x)g(y)

what is direct integration?

Thanks

- #4

Tide

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[tex]\int_{x_0}^{x} \frac {d}{dx'}\left(x'^2 y\right) dx' = \int_{x_0}^{x} \frac {x'^2}{1+x'^2} dx'[/tex]

- #5

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

- #6

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[tex] \int du = u +\mathcal{C} [/tex]

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

Daniel.

- #7

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okey, I understand what you just wrote :P

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

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

- #8

HallsofIvy

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sony said:

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 x

"Direct integration" just meant regular integration!

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

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- #9

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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

HallsofIvy

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Oops! Sorry about that. I misread the original equation!sony said:

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...)

It's not that the yx

Given

[tex]x^2y= \int d(x^2y)= \int \frac{x^2dx}{x^2+ 1}[/tex].

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]

and it becomes easy to integrate.

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- #11

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Oh, thanks! Now I see it :)

- #12

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

- #13

HallsofIvy

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

Fundamental Theorem of Calculus!

[tex] \int d(Anything)= Anything+ C[/tex]

[tex]\int d(x^2y)= x^2y + C[/tex]

- #14

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Oh! sorry! I see it! bah

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

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

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