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

cwrw238
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3exp (ln(1x^2)^(1/2))= sqrt(1x^2)

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

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

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

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

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

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

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

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

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

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

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

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

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