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cwrw238
Group Title
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
 one year ago
cwrw238 Group Title
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
 one year ago

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

phi Group TitleBest ResponseYou've already chosen the best response.3
that looks good
 one year ago

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

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

phi Group TitleBest ResponseYou've already chosen the best response.3
I 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 \[ y= \frac{A}{\sqrt{1x^2}}  1\]
 one year ago

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

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

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

phi Group TitleBest 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 \[ \frac{1}{2} \int u^{ \frac{1}{2}} du=  \frac{1}{2} \cdot 2 u^{\frac{1}{2}} \]
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
oh 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)
 one year ago

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

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

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

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
yep 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) ?
 one year ago

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

phi Group TitleBest ResponseYou've already chosen the best response.3
The 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
 one year ago

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

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
yes 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
 one year ago

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