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 one year ago
I'm having trouble solving this differential equation
(1  x^2) y'  xy = x
I'm not sure about the integrating factor
is it (1  x^2) ^(1/2) or 1/ ((1  x^2)^(1/2)
(or maybe its neither!). Please help.
 one year ago
I'm having trouble solving this differential equation (1  x^2) y'  xy = x I'm not sure about the integrating factor is it (1  x^2) ^(1/2) or 1/ ((1  x^2)^(1/2) (or maybe its neither!). Please help.

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phi
 one year ago
Best ResponseYou've already chosen the best response.3if in the form y' + P(x) y = Q(x) the integrating factor is e^F where F= integral P(x) dx

SmartWish
 one year ago
Best ResponseYou've already chosen the best response.0\[(1x^2)\frac{dy}{dx} xy = x\]

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1@ph1  yes  I remembered that I'm (to my shame) confuse by the algebra

phi
 one year ago
Best ResponseYou've already chosen the best response.3integral x dx/(1x^2) = 1/2 * ln(1x^2) (toss in abs signs to be sure it's positive)

phi
 one year ago
Best ResponseYou've already chosen the best response.3exp (ln(1x^2)^(1/2))= sqrt(1x^2)

phi
 one year ago
Best ResponseYou've already chosen the best response.3integral du/u = ln(u) in this case u = 1x^2 du = 2x dx

phi
 one year ago
Best ResponseYou've already chosen the best response.3\[ e^{ab} = \left(e^a\right)^b \] so \[ e^{\ln(1x^2)\cdot \frac{1}{2}}=\left(e^{\ln(1x^2)}\right)^\frac{1}{2} \] \[= (1x^2)^\frac{1}{2}= \sqrt{1x^2} \]

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1ok  i see it now . I was confused with the sign when integrating thanx

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1i need more practice in FODEs .......

phi
 one year ago
Best ResponseYou've already chosen the best response.3are you saying you can't finish ?

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1no i think i've got it now I get the solution to be y = A(1  x^2)^(1/2)  1

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1I got it down to dw:1359820446243:dw

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1from here its easy thanks again

phi
 one year ago
Best ResponseYou've already chosen the best response.3I would expect to see \[\sqrt{1x^2}y = \int\limits \sqrt{1x^2}\frac{x}{1x^2} dx\] the integral gives \[  \sqrt{1x^2}+ A\] so \[\sqrt{1x^2}y = A  \sqrt{1x^2} \] and finally \[ y= \frac{A}{\sqrt{1x^2}}  1\]

phi
 one year ago
Best ResponseYou've already chosen the best response.3though you could have tossed in a minus sign on both sides before integrating ?

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1hmm i'm confused how come the integral is negative of square root bit?

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1i'm allergic to integration!!!

phi
 one year ago
Best ResponseYou've already chosen the best response.3\[\int\limits \sqrt{1x^2}\frac{x}{1x^2} dx = \int (1x^2)^{ \frac{1}{2}} x dx\] u is 1x^2 and du = 2x dx we need a 2 so multiply and divide by 2: \[ \frac{1}{2} \int (1x^2)^{ \frac{1}{2}} 2x dx\] we now have \[ \frac{1}{2} \int u^{ \frac{1}{2}} du=  \frac{1}{2} \cdot 2 u^{\frac{1}{2}} \]

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1oh ok  I used the following standard form which probably isnt valid in this case Integral x dx / sqrt(a^2 + x^2) = sqrt (a^2 + x^2)

phi
 one year ago
Best ResponseYou've already chosen the best response.3except you have a^2  x^2 not a^2 + x^2

phi
 one year ago
Best ResponseYou've already chosen the best response.3maybe the Greeks had the right idea, stick with positive rational numbers at all times

phi
 one year ago
Best ResponseYou've already chosen the best response.3then none of these stray minus signs to keep track of.

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1yep this is probably a silly question but since the left hand side is the result of the product rule shoudntt it be  y sqrt(1  x^2) ?

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1yes its the signs which have confused me here

phi
 one year ago
Best ResponseYou've already chosen the best response.3The integrating factor is sqrt(1x^2). It is positive. call the integrating factor F(x) and start with dy/dx + P(x) y = Q(x) F(x) * dy/dx + F(x)*P(x) y = F(x)*Q(x) and F(x) dy + P(x) y F(x) dx = F(x)Q(x) dx remember F(x) = exp(integral(P(x))) d F(x) = P(x) exp( integral(P(x))) dx we could work through the details, but it looks ok to me. we hae F(x) dy + y d (F(x)) = F(x)Q(x) dx or d( F(x) y) = F(x)Q(x) dx

phi
 one year ago
Best ResponseYou've already chosen the best response.3the "tricky part" is taking the derivative of \[ \frac{d}{dx} e^{\int P(x) dx} = e^{\int P(x) dx} \cdot \frac{d}{dx}\int P(x) dx= e^{\int P(x) dx} \cdot P(x) dx\]

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1yes ok i'll have a good look at this  I'm still a bit mistifi ed (maybe its because i've got a cold!) ty

phi
 one year ago
Best ResponseYou've already chosen the best response.3if this was easy, anybody could do it.
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