# Differential Equations

1. Jan 15, 2007

### The Bob

Hey all,

I have an exam coming up and the pattern of questions seems to include some that I am stumped with.

One goes like this:

The second order differential equation

x2y'' - 6xy' + 12y = 0,​

has non-constant coefficients. Find the general solution by looking for a solution y = xn and finding a quadratic equation for n.

I would love to say "this is what I have done" but I just couldn't see anything. I tried substituting y = xn, y' = nxn-1 and y'' = n(n-1)xn-2 into the equation but that just doesn't help, so far as I can see. I need a nudge in the right direction.

Cheers,

Last edited: Jan 15, 2007
2. Jan 15, 2007

### cristo

Staff Emeritus
What do you get when you substitute y=xn (and derivatives) into the equation? Why do you say that this won't help? I think that's the correct way to approach this question!

Last edited: Jan 15, 2007
3. Jan 15, 2007

### Integral

Staff Emeritus
Show us what you get after making the substitutuons for

$$y = x^n$$ and its derivatives.

4. Jan 15, 2007

### The Bob

Ok, well... after substitution and messing around with powers, which is probably why I didn't see it going anywhere (at least), I get:

xn(n - 4)(n - 3) = 0​

I am assuming I would be correct to say that n = 4 and n = 3.

So y = x3 + x4 with the quadratic for n as (n - 4)(n - 3) = 0

How much of this is good and how much can you tell is guess work?

Cheers,

Last edited: Jan 15, 2007
5. Jan 15, 2007

### morson

Wouldn't you just leave it as "y = x^4, y = x^3"?

6. Jan 16, 2007

It's all correct. When you sub in y = x^n, since the differential equation has the form $ax^2y'' + by' + cy = 0$, the reduction in the power due to differentiations on $x^n$ are cancelled out by the x^2 in front of y" and the x in front of y'. You can thus factor out the $x^n$ to get $$x^n(an(n-1) + bn + c) = 0[/itex], which you know is only true for all possible values of x if the quadratic in n is equal to zero. In the specific problem you mentioned, that's why n = 3, 4. And so, you get the solutions y = x^4 and y = x^3, and your general solution is a linear combination of them, $y = Ax^3 + Bx^4$. So it appears you knew what you were doing, you just weren't sure of it. 7. Jan 16, 2007 ### HallsofIvy Staff Emeritus You don't need to "assume" it. This equation must be true for all x. In particular for x= 1 which means (n-4)(n-3)= 0. Youwere good up to this point! Any solution to a second order, linear, homogeneous differential equation can be written as a linear combination of two independent solutions. x3 and x4 are independent solutions. What does a linear combination of them look like? 8. Jan 16, 2007 ### The Bob Yeap, this entire step eluded me. Well it didn't but and it is annoying because when I looked at it first time I thought about doing somthing like this. y = Ax3 + Bx4, although I have worked it through and it would appear that A and B are can be any constant. Anyway cheers all, The Bob (2004 ©) 9. Jan 16, 2007 ### The Bob Whilst I am on a role, I have another question before tomorrow. Given that y = ex is one solution of the differential equation (1 + x)y'' - (1 - 2x)y' + (x - 2)y = 0​ find the general solution. What I have done is let y = uex, differentiate this and substitued in. Then I have let w = u' and got down to: w'(ex + xex) + 3exw = 0​ The problem from here is that the way I have been told to solve it is to assume it is in the form: y' + P(x)y = Q(x)​ which does not allow a solution, unless, again, I am doing it wrong. Any pointers? Cheers, The Bob (2004 ©) 10. Jan 17, 2007 ### Mute Keep in mind that e^x is never zero, so you can divide it out of your equation in w to get [tex](1+x)w' + 3w = 0$$,

which is a linear first order differential equation that you can solve using an integrating factor.

11. Jan 17, 2007

### dextercioby

I'd say it's better if you simply separate variables and integrate instead of calculating the integrating factor.

Daniel.

12. Jan 17, 2007

### The Bob

May I thank you both. Although I do not understand completely why you can simply divide through by ex because it is never zero, I can appreciate what is being said and, unfortunately for now, learn that bit parrot fashion.

After this division, I could see where the seperable equation was with more ease.

Cheers again

The Bob (2004 )

13. Jan 17, 2007

### HallsofIvy

Staff Emeritus
One serious problem you have with this is that ex is NOT a solution to that equation!

Apparently, from what you have below the equation is actually
(1+ x)y"+ (1- 2x)y'+ (x-2)y= 0.

14. Jan 17, 2007

### Mute

What if instead of e^x you had had $(2+2x)w' + 6w = 0$ or $(117 + 117x)w' + 351w = 0$? In each case you would multiply both sides of the equation by 1/2 and 1/117, respectively, to get rid of the common factor in each term. The same goes for e^x: You're multiplying both sides of the equation by 1/e^x = e^(-x) to remove those terms from your equation. Equivalently, you could just factor it out, like you do with the $x^n$ in your first problem, to get $e^x((1+x)w' + 3w) = 0$ and note that for this equation to hold for all x, you need $(1+x)w' + 3w = 0$.

Last edited: Jan 17, 2007