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ODE, bernoulli equation -> leads to crazy integral !

  • Thread starter Jonnyb42
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ODE, bernoulli equation --> leads to crazy integral !

Homework Statement



An Initial Value Problem, ODE (Bernoulli equation)

ODE: [x^2]*y' + 2*[x^3]*y = [y^2]*(1+2*[x^2])
IV: y(1) = 1/2

Homework Equations



general form of Bernoulli's equation:
y' + a(x)y = b(x)*[y^n]

First order, linear ODE form:
y' + a(x)y = b(x)

The Attempt at a Solution



I begin by simplifying the Bernoulli-type equation into a linear form, at which point I end up with an integral that I DO NOT KNOW HOW TO SOLVE.

9k85f6.jpg


I have either really messed up somewhere along the way to that Integral, or not and I just dont' know how to solve it.

Thanks for any help.
 

Answers and Replies

  • #2
tiny-tim
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Hi Jonnyb42! :smile:

Looks ok to me …

but you can't solve that integral unless you're allowed to use the "error function" erf(x). :redface:
 
  • #3
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No further ideas here, sorry :-\
 
  • #4
HallsofIvy
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Are you required to actually integrate it? I remember in my first D.E. class, I had one problem in the text that I could reduce to an integral but could not integrate. If finally gave up and looked in the back of the book where I found that the answer was given in terms of that integral!
 
  • #5
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But how could I get it even in terms of the integral? I need a constant of integration to solve for so y(1) = 1/2
 
  • #6
ehild
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Substitute z=x/y. You will get an easy separable differential equation.

ehild
 
Last edited:
  • #7
ehild
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With the substitution z=x/y,

y=x/z, y'=(z-xz')/z^2,the ODE becomes

x^2(z-xz')/z^2+2x^4/z=(1+2x^2)x^2/z^2.

Multiplying with z^2/x^2, and rearranging the equation:

z-xz'+2x^2 z=(1+2x^2), which is a separable ODE:

xz'=(1+2x^2)(z-1)--->z'/(z-1)=(1+2x^2)/x

ehild
 
  • #8
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ehild, thank you!!

How do you know to do that substitution?

thanks again,
Jonny
 
  • #9
ehild
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I did not know that substitution, I just tried...

Dividing both sides of the equation by y^2*x^2, it becomes

y'/y^2+ 2*(x/y) = 1/x^2+2.

So why not try to choose x/y the new variable? The derivative of x/y will contain the term y'/y^2...

There are no recipes for every ODE-s, and there can be different methods for the same ODE. The recipe suggested the substitution z=1/y, but it leads to a difficult integral. You have to be creative :smile:

ehild
 
  • #10
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thanks very much!!!!!!!!

jonny
 
  • #11
tiny-tim
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oops!

Hi Jonnyb42! :smile:

Looks ok to me …

but you can't solve that integral unless you're allowed to use the "error function" erf(x). :redface:
oops! :redface:

your original method was fine :smile:, and on careful inspection your final integral

∫ -e-x2(2 + 1/x2) dx​

is simply e-x2/x + C :blushing:
 
  • #12
ehild
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your original method was fine :smile:, and on careful inspection your final integral

∫ -e-x2(2 + 1/x2) dx​

is simply e-x2/x + C :blushing:
Ingenious! How can one find out this integral?

ehild
 
  • #13
tiny-tim
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just by looking hard enough …

if it's (fe-x2)',

then that's (f' -2xf)e-x2, so -2xf = -2 works. ie f = 1/x, f' = -1/x2 :wink:
 
  • #14
ehild
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You can look really hard. :smile:
Those who do not have such eyes would input it to Wolframalpha...

ehild
 
  • #15
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It's an initial value problem. So, if you have a linear ODE:
[tex]
z' + a(x) z = b(x)
[/tex]
and you find the integrating factor:
[tex]
\mu(x) = e^{\int{a(x) \, dx}}
[/tex]
without caring about the arbitrary integrating constant, because that would give an arbitrary multiplicative factor to your integrating factor, then, the solution to the initial value problem with:
[tex]
z(x_{0}) = z_{0}
[/tex]
is:
[tex]
\mu(x) \, z(x) - \mu(x_{0}) \, z_{0} = \int_{x_{0}}^{x}{\mu(t) \, b(t) \, dt}
[/tex]

From here you can immediately see that if [itex]\mu(x)[/itex] gets multiplied by an arbitrary constant, each term in the above expression gets multiplied by it and the equality holds.
 

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