Solving differential equations

It would have been better to write it as:f(3) = e^{[\ln(\sqrt{10})+\ln(5/sqrt{5})]} = e^{[\ln(\sqrt{10})+\ln(5)-\ln(\sqrt{5})]} = e^{\ln(\sqrt{10})} \cdot e^{\ln(5)} \cdot e^{-\ln(\sqrt{5})} = \sqrt{10} \cdot 5 \cdot \frac{1}{\sqrt{5}} = 5 \sqrt{2}.Yeah, to be honest, I already thought that step was too big. :redface: It would have been better to write it as
  • #1
Shannabel
74
0

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?
 
Physics news on Phys.org
  • #2
If [itex]u = x^2[/itex], then [itex]du = 2 x dx[/itex].
So instead of
[tex]\ln y = 2(u+1)/du [/tex]
you get
[tex]d(\ln y) = du/[2(u + 1)] [/tex]
and, integrating,
[tex]\ln y = \ln[ 2(u + 1) ] + c[/tex].
 
Last edited:
  • #3
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
 
  • #4
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?
 
  • #5
@CompuChip: you forgot a factor (1/2) in your last line...
 
  • #6
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:
[tex]5 = e^{[\ln \sqrt 5+c]}[/tex]
You'll see that you'll get a different value for c. :smile:
 
  • #7
I like Serena said:
You forgot a couple of brackets...

You should use:
[tex]5 = e^{[\ln \sqrt 5+c]}[/tex]
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?
 
  • #8
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.
[tex]5 = e^{(\ln \sqrt 5)+c}[/tex]
Here you need to simplify by applying the rules for powers.
That is [itex]e^{a+b}=e^a \cdot e^b[/itex].
 
  • #9
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.
[tex]5 = e^{(\ln \sqrt 5)+c}[/tex]
Here you need to simplify by applying the rules for powers.
That is [itex]e^{a+b}=e^a \cdot e^b[/itex].

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:
[tex]y = e^{(1/2)\ln(x^2+1)+c} [/tex]
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:
[tex]y = e^{(1/2)\ln(x^2+1)+c} [/tex]
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:
 

FAQ: Solving differential equations

1. What is a differential equation?

A differential equation is a mathematical equation that relates an unknown function to its derivatives. It is used to describe the relationship between a quantity and its rate of change.

2. Why are differential equations important?

Differential equations are important because they are used to model and solve many real-world problems in fields such as physics, engineering, economics, and biology. They allow us to predict the behavior of a system over time and make informed decisions based on those predictions.

3. What are the different methods for solving differential equations?

There are several methods for solving differential equations, including separation of variables, substitution, and using integrating factors. Other methods include power series, Laplace transforms, and numerical methods such as Euler's method and Runge-Kutta methods.

4. How do I know which method to use for a specific differential equation?

The method used to solve a differential equation depends on its type and order. For example, first-order differential equations can often be solved using separation of variables, while higher-order equations may require more advanced methods such as Laplace transforms. It is important to carefully analyze the equation and choose the appropriate method accordingly.

5. Can differential equations be solved analytically or only numerically?

Both analytical and numerical methods can be used to solve differential equations. Analytical solutions involve finding an explicit expression for the solution, while numerical solutions involve approximating the solution using numerical algorithms. In some cases, an analytical solution may not be possible and a numerical approach is necessary.

Back
Top