Solving differential equations

Shannabel
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Homework Statement


If (x^2+1)dy/dx = xy for all y>0, and y(2) = 5, then y(3) = ?


Homework Equations





The Attempt at a Solution


dy/y = xdx/(x^2+1) -- use substitution with u=x^2
lny = 2(u+1)/du
lny = 2ln(x^2+1)+c
y = e^(2ln(x^2+1)+c)
5 = e^(2ln(5)+c)
5 = e^(ln25+c)
5 = 25+c
c = 20

y(3) = e^(2ln10+20)
= e^ln100+20
= 100+20
= 120



... the answer should be 5sqrt(2), so I am very off, but I can't find why. help?
 
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If u = x^2, then du = 2 x dx.
So instead of
\ln y = 2(u+1)/du
you get
d(\ln y) = du/[2(u + 1)]
and, integrating,
\ln y = \ln[ 2(u + 1) ] + c.
 
Last edited:
We have exp[(1/2)*ln(x^2+ 1) + c] = k*sqrt(x^2+1), where k =exp(c). Note: x*dx/(1+x^2) = (1/2)du/(u+1).

RGV
 
Ray Vickson said:
We have exp[(1/2)*ln(x^2+ 1) + c] = k*sqrt(x^2+1), where k =exp(c). Note: x*dx/(1+x^2) = (1/2)du/(u+1).

RGV

if y = e^(1/2)ln(x^2+1)+c
5 = e^lnsqrt(5)+c
then i end up with c = 5-sqrt(5), and then
y(3) = e^1/2ln10+5-sqrt(5)
= sqrt(10)+5-sqrt(5)
... confused?
 
@CompuChip: you forgot a factor (1/2) in your last line...
 
Shannabel said:
if y = e^(1/2)ln(x^2+1)+c
5 = e^lnsqrt(5)+c
then i end up with c = 5-sqrt(5), and then
y(3) = e^1/2ln10+5-sqrt(5)
= sqrt(10)+5-sqrt(5)
... confused?

You forgot a couple of brackets...

You should use:
5 = e^{[\ln \sqrt 5+c]}
You'll see that you'll get a different value for c. :smile:
 
I like Serena said:
You forgot a couple of brackets...

You should use:
5 = e^{[\ln \sqrt 5+c]}
You'll see that you'll get a different value for c. :smile:

do you mean then i should get
5 = sqrt5 +e^c?
how do i simplify that?
 
Shannabel said:
do you mean then i should get
5 = sqrt5 +e^c?
how do i simplify that?

No, that is not right.
Let's take it one step back.
Let me put the braces slightly diffently for clarity.
5 = e^{(\ln \sqrt 5)+c}
Here you need to simplify by applying the rules for powers.
That is e^{a+b}=e^a \cdot e^b.
 
I like Serena said:
No, that is not right.
Let's take it one step back.
Let me put the braces slightly diffently for clarity.
5 = e^{(\ln \sqrt 5)+c}
Here you need to simplify by applying the rules for powers.
That is e^{a+b}=e^a \cdot e^b.

so then don't i get
5 = e^ln(sqrt5) + e^c?
and e^ln(sqrt5) = sqrt5?
 
  • #10
Shannabel said:
so then don't i get e^ln(sqrt5) = sqrt5?

Yes! :smile:


Shannabel said:
5 = e^ln(sqrt5) + e^c?

No. :frown:
 
  • #11
I like Serena said:
Yes! :smile:




No. :frown:

oh, durrr
5 = sqrt5*e^c?
 
  • #12
Shannabel said:
oh, durrr
5 = sqrt5*e^c?

Yes! :wink:
 
  • #13
I like Serena said:
Yes! :wink:

YAY!
so from there, i get
5/sqrt5 = e^c?
so c = ln(5/sqrt5)?
 
  • #14
Yes, so... :rolleyes:
 
  • #15
I like Serena said:
Yes, so... :rolleyes:

hahaha
f(3) = e^lnsqrt10+ln5/sqrt5
= sqrt10+5/sqrt5

... annnd I'm off track again lol
 
  • #16
Err... can you do the "oh, durrr" thing again? :shy:
 
  • #17
I like Serena said:
Err... can you do the "oh, durrr" thing again? :shy:

well that was dumb.
e^lnsqrt10+ln5/sqrt5 = e^lnsqrt10*e^ln5/sqrt5
= sqrt10*5/sqrt5

I GOT IT YAYYYY :):)
thanksyou :)
 
  • #18
Congratulations! :biggrin:One thing though...

You started with:
y = e^{(1/2)\ln(x^2+1)+c}
But that is not the equation you had in your first post... o:)

How did you get it?
 
  • #19
I like Serena said:
Congratulations! :biggrin:


One thing though...

You started with:
y = e^{(1/2)\ln(x^2+1)+c}
But that is not the equation you had in your first post... o:)

How did you get it?

:D

i changed the first equation to
dy/y = xdx/(x^2+1), then used the substitution u=x^2 on the right side :)
 
  • #20
Shannabel said:
:D

i changed the first equation to
dy/y = xdx/(x^2+1), then used the substitution u=x^2 on the right side :)

Yeah, I got that...
So did you follow CompuChip's correction and then my correction on CompuChip's correction? :confused:
 
  • #21
I like Serena said:
Yeah, I got that...
So did you follow CompuChip's correction and then my correction on CompuChip's correction? :confused:

i was confused by what you were saying but i did it over and realized i should have had 1/2 instead of 2 :)
 
  • #22
Shannabel said:
i was confused by what you were saying but i did it over and realized i should have had 1/2 instead of 2 :)

Oh, that's okay then! :approve:
 
  • #23
Shannabel said:
hahaha
f(3) = e^lnsqrt10+ln5/sqrt5
= sqrt10+5/sqrt5

... annnd I'm off track again lol

Why don't you read my previous response?

RGV
 
  • #24
Ray Vickson said:
Why don't you read my previous response?

RGV

i did, it confused me :(
 
  • #25
Ray Vickson said:
Why don't you read my previous response?

RGV
Shannabel said:
i did, it confused me :(
Yeah, to be honest, I already thought that step was too big. :redface:
 
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