A faster way to solve this ODE?

  • #1

Main Question or Discussion Point

[tex]\frac{dy}{dx}[/tex][x^3-12y]=x^2

That ODE took me a little while to solve... I had to switch from dy/dx to dx/dy, do a Bernoulli substitution, two applications of integration by parts, and then some algebraic manipulations in order to solve it.

The final result?
y = (x^3-4)/12

This really simple equation which satisfies the ODE leads me to believe that there is an easier method for solving this... A few of these might appear on my final tomorrow, and I don't want to spend all of that time if there is an easier method.

Thanks!
 

Answers and Replies

  • #2
Hootenanny
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Whilst this reply will come a little too late for your final, it may still be useful. Notice that if we assume that [itex]x\neq0[/itex] we may write,

[tex]\frac{dy}{dx} - \frac{12}{x^3}y=\frac{1}{x}[/tex]

Which is a much easier to solve.
 
  • #3
Defennder
Homework Helper
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What happened to the dy/dx term in your [tex]- \frac{12}{x^3} y[/tex]?
 
  • #4
Ya... The derivative was supposed to be distributed.
 
  • #5
Looking at the de, if you were to power expand y (heck you could even Laurent expand it) in order to preserve the right hand side y would have to be cubic, that is the powers can't be less than 0 nor can they be greater than 3.

So say there exists a,b,c,d such that [tex]y = ax^3 + bx^2 + cx + d[/tex] then we have

[tex]\frac{dy}{dx}(x^3-12y) = (3 ax^2 + 2bx + c)[(1 - 12a)x^3 -12bx^2-12cx-12d][/tex]

but that has to equal x^2 so the x^5 term must go away
[tex]
3a(1-12a)=0
[/tex]
and clearly a is not zero so [tex]a=1/12[/tex].

The x^4 term must also go away
[tex]
-12b(3a) = 0
[/tex]
and we know a is not zero so that means that [tex]b=0[/tex].

Let's restate the equation again
[tex]
(\frac{1}{4}x^2+c)(-12cx-12d)=x^2
[/tex]

The x^3 term goes away so
[tex]
3c=0[/tex]
or [tex]c=0[/tex].

Alright finally the x^2 term should have a coefficient of 1 so
[tex]
-3d=1
[/tex]
or [tex]d=-1/3[/tex].

And we have the solution [tex]y=\frac{1}{12}x^3-\frac{1}{3}[/tex]

This is not a simpler method to solve it, but I think it shows more clearly why the solution is so simple.
 
  • #6
tiny-tim
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[tex]\frac{dy}{dx}[/tex][x^3-12y]=x^2
Hi scorpion990! :smile:

The way I'd do it is to assume that y is a polynomial in x.

If its highest term is x^n, then you start with nx^(n-1)[x^3 - 12x^n], so the highest terms are 3+n-1 and/or n+n-1, and so … :smile:
 
  • #7
Hootenanny
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What happened to the dy/dx term in your [tex]- \frac{12}{x^3} y[/tex]?
Sorry guys, I misread the parenthesise, my bad.
 
  • #8
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Make it exact?

Why wouldn't you make it exact first.

[tex] (x^3 - 12y)dy + (-x^2)dx = 0 [/tex]

By inspection one can find [itex]e^\left-3y\right[/itex] makes this exact.

[tex] e^\left-3y\right\right((x^3 - 12y)dy + e^\left-3y\right\right((-x^2)dx = 0 [/tex]
[tex] \smallint e^\left-3y\right\right((x^3 - 12y)dy [/tex]
[tex] \smallint e^\left-3y\right\right((-x^2)dx [/tex]

After the one integration by parts that is required, you end up with:
[tex] C = \frac{e^\left-3y\right}{3}(4 - x^3 + 12y) [/tex]

I guess you have an initial condition that tells you that C is zero.

This may or may not be considered quicker, but it was the only way I remembered after taking DE 10+ years ago. :)

Keep these problems coming. I'm going to go back to school and get a master's, assuming the wife lets me. :)
 
  • #9
Defennder
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Well I certainly couldn't see that would have worked! I mean, how did you know that some integrating factor existed which could make it exact?
 
  • #10
D H
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After the one integration by parts that is required, you end up with:
[tex] C = \frac{e^\left-3y\right}{3}(4 - x^3 + 12y) [/tex]
Nice work. Everyone else has found a particular solution. You found the general solution.
 
  • #11
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When I took DE in college, I would always look to make it exact first. Since the @/@y of the second term is zero and the @/@x of the first term is in terms of x only, you know that the integrating factor is probably a function of y only. You end up with

[tex] F(y) = -F'(y)/3 [/tex]
 

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