A faster way to solve this ODE?

[scorpion990]

$$\frac{dy}{dx}$$[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!

 

[Hootenanny]

Whilst this reply will come a little too late for your final, it may still be useful. Notice that if we assume that $x\neq0$ we may write,

$$\frac{dy}{dx} - \frac{12}{x^3}y=\frac{1}{x}$$

Which is a much easier to solve.

 

[Defennder]

What happened to the dy/dx term in your $$- \frac{12}{x^3} y$$?

 

[scorpion990]

Ya... The derivative was supposed to be distributed.

 

[DavidWhitbeck]

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 $$y = ax^3 + bx^2 + cx + d$$ then we have

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

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

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

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

The x^3 term goes away so
$$3c=0$$
or $$c=0$$.

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

And we have the solution $$y=\frac{1}{12}x^3-\frac{1}{3}$$

This is not a simpler method to solve it, but I think it shows more clearly why the solution is so simple.

 

[tiny-tim]

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

Hi scorpion990!

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 …

 

[Hootenanny]

What happened to the dy/dx term in your $$- \frac{12}{x^3} y$$?

Sorry guys, I misread the parenthesise, my bad.

 

[sennyk]

Make it exact?

Why wouldn't you make it exact first.

$$(x^3 - 12y)dy + (-x^2)dx = 0$$

By inspection one can find $e^\left-3y\right$ makes this exact.

$$e^\left-3y\right\right((x^3 - 12y)dy + e^\left-3y\right\right((-x^2)dx = 0$$
$$\smallint e^\left-3y\right\right((x^3 - 12y)dy$$
$$\smallint e^\left-3y\right\right((-x^2)dx$$

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

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 let's me. :)

 

[Defennder]

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?

 

[D H]

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

Nice work. Everyone else has found a particular solution. You found the general solution.

 

[sennyk]

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

$$F(y) = -F'(y)/3$$