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First Order Differential Equations

  • Thread starter Bucky
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Homework Statement



Solve the following differential equation using seperation of variables

[tex] (1+x)^2 y' = (1-y)^2 , y(1) = 2 [/tex]

Homework Equations





The Attempt at a Solution


haven't gotten very far in this at all :/

i've tried dividing both sides by [tex](1+x)^2[/tex], in order to get y' on it's own..

[tex]y' = \frac{(1-y)^2}{(1+x)^2} [/tex]


but i don't know how to integrate this...but had a go anyway

apparently the rule for integrating an expression in brackets is..

[tex] \frac{(ax + b)^n}{a(n+1)} [/tex]

so i tried integrating both halves of the fraction seperatley...giving
[tex]
\frac{(-y+1)^3 }{-3y}[/tex]
[tex]
\frac{(x+1)^2}{3x}
[/tex]
putting these together and dividing gave

[tex]\frac{3x(-y+1)^2}{-3y(x+1)^3} + C[/tex]


however i don't think this is accurage, as substituting in 1 and 2 for x and y respectivley gmade C some out as -3/48. Can someone shed some light at where I've went wrong?
 
Last edited:

Answers and Replies

  • #2
Dick
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You haven't gotten y on it's own. Write y'=dy/dx and put the dy with the y's and the dx with the x's.
 
  • #3
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are you saying i need to get the y terms on one side of the equals and the x terms on the other side?

if so then i'm stumped. I'm not sure how to split up the equation in it's initial form.
 
  • #4
Dick
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How about dy/(1-y)^2=dx/(1+x)^2? Does it look split now?
 
  • #5
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ok, right, rearranged it and heres what i have:

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

[tex]\frac {1}{(1-y)^3}{3} = {1}{(1+x)^3}{3} + C[/tex]

which, after substituting y(0) = -1 (wrong values on first post BTW) i get

[tex]\frac{-7}{24} = C[/tex]

I'm pretty sure this is wrong...but anyway lets sub that back into the integrated function

[tex] \frac{1}{3} (1-y)^3 = \frac{1}{3} (1+x)^3 - 7/24 [/tex]
[tex] \frac 8(1-y)^3 = 8(1+x)^3 - 7[/tex]

the answer in teh book is given as

[tex] y = \frac{x-1}{1+3x}[/tex]
 
  • #6
Dick
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It would help a lot if you did the integrations correctly. The integral of 1/x^2 is not 1/x^3.
 
  • #7
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er sorry thats a typo, i get the integral of 1/(1+x)^3 to be....

1/3(1+x)^3
 
  • #8
Dick
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You want to integrate 1/(1+x)^2. Try again.
 
  • #9
Dick
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Hint: integral of 1/u^2 du=-1/u.
 
  • #10
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so...the integral of 1/(1+x)^2 is just -1 - (1/x) ?


even that doesn't seem right...though my formula book doesn't have the formula for such an instance.
 
  • #11
Dick
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Make a substitution u=x+1. Similarly for the y integral. You have done that, right?
 
  • #12
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ok so that gives...

[tex]- \frac{1}{(1-y)} = - \frac{1}{(1+x)} [/tex]
 
  • #13
HallsofIvy
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ok, right, rearranged it and heres what i have:

[tex]\int \frac{dy}{(1-y)^2} = \int \frac{dx}{(1+x)^2}[/tex]
I added the integral sign on the right side.

[tex]\frac {1}{(1-y)^3}{3} = {1}{(1+x)^3}{3} + C[/tex]
I think you meant
[tex]\frac{1}{3(1-y)^3}= \frac{1}{3(1+x)^3}+ C[/tex]
but you've missed a sign. But only one sign. As Dick said you need to substitute for both 1+ x and 1- y. If u= 1- y, what is du?
 
  • #14
Dick
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I think you meant
[tex]\frac{1}{3(1-y)^3}= \frac{1}{3(1+x)^3}+ C[/tex]
but you've missed a sign. But only one sign. As Dick said you need to substitute for both 1+ x and 1- y. If u= 1- y, what is du?
I think Halls was just cleaning up your notation and doesn't mean to imply that cube is the right power. But you did miss the substitution sign.
 
  • #15
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i don't see where the missing sign is. unless the sign attatched to the letter comes out? but thats not what i thought you meant when you said

1/u^2 = -1/u

so do you mean that i should have...

[tex]\frac{dy}{(1-y)^2} = \frac{dx}{(1+x)^2}[/tex]
[tex]-\frac{dy}{(1-y)^2} = \frac{dx}{(1+x)^2}[/tex]

EDIT: wait that can't be right...yeah i'm lost.
 
Last edited:
  • #16
Dick
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ok so that gives...

[tex]- \frac{1}{(1-y)} = - \frac{1}{(1+x)} [/tex]
If you differentiate the right side of this you get 1/(1+x)^2. That's what you want. If you differentiate the left side you get -1/(1-y)^2. Don't forget the chain rule. That's NOT what you want.
 
  • #17
HallsofIvy
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i don't see where the missing sign is. unless the sign attatched to the letter comes out? but thats not what i thought you meant when you said

1/u^2 = -1/u
He certainly never said that! he said that the anti-derivative of 1/u2 is -1/u.

so do you mean that i should have...

[tex]\frac{dy}{(1-y)^2} = \frac{dx}{(1+x)^2}[/tex]
[tex]-\frac{dy}{(1-y)^2} = \frac{dx}{(1+x)^2}[/tex]

EDIT: wait that can't be right...yeah i'm lost.
If you let u= 1+ x, then du= dx and dx/(1+x)2= du/u2. What's the anti-derivative of that?

If you let v= 1-y, then dv= -dy and y/(1-y)2= -dv/v2. What's the anti-derivative of that?
 
  • #18
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ok so i tried a totally different method and it seems to work.

instaed of substitution i tried using the standard integral

[tex] \int (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)} [/tex]

thanks a lot for your help guys!
 
  • #19
Dick
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ok so i tried a totally different method and it seems to work.

instaed of substitution i tried using the standard integral

[tex] \int (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)} [/tex]

thanks a lot for your help guys!
Let u=(ax+b), du=a*dx, du/a=dx.

[tex] \int (ax+b)^n dx= \frac{1}{a} \int (u)^n du = \frac{1}{a}
\frac{u^(n+1)}{(n+1)} = \frac{(ax+b)^{n+1}}{a(n+1)}[/tex]

That is a TOTALLY DIFFERENT METHOD!
 
  • #20
Dick
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ok so i tried a totally different method and it seems to work.

instaed of substitution i tried using the standard integral

[tex] \int (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)} [/tex]

thanks a lot for your help guys!
Let u=(ax+b), du=a*dx, du/a=dx.

[tex] \int (ax+b)^n dx= \frac{1}{a} \int (u)^n du = \frac{1}{a}
\frac{u^{n+1}}{(n+1)} = \frac{(ax+b)^{n+1}}{a(n+1)}[/tex]

That is a TOTALLY DIFFERENT METHOD!
 

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