• Support PF! Buy your school textbooks, materials and every day products Here!

Differential Equations

  • Thread starter The Bob
  • Start date
1,100
0
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,

The Bob (2004 ©)
 
Last edited:

Answers and Replies

cristo
Staff Emeritus
Science Advisor
8,056
72
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:
Integral
Staff Emeritus
Science Advisor
Gold Member
7,184
55
Show us what you get after making the substitutuons for

[tex] y = x^n [/tex] and its derivatives.
 
1,100
0
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!
Show us what you get after making the substitutuons for

[tex] y = x^n [/tex] and its derivatives.
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,

The Bob (2004 ©)
 
Last edited:
35
0
Wouldn't you just leave it as "y = x^4, y = x^3"?
 
Mute
Homework Helper
1,381
10
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,

The Bob (2004 ©)
It's all correct. When you sub in y = x^n, since the differential equation has the form [itex]ax^2y'' + by' + cy = 0[/itex], the reduction in the power due to differentiations on [itex]x^n[/itex] are cancelled out by the x^2 in front of y" and the x in front of y'. You can thus factor out the [itex]x^n[/itex] to get [tex]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, [itex]y = Ax^3 + Bx^4[/itex].

So it appears you knew what you were doing, you just weren't sure of it.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
899
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.
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.

So y = x3 + x4
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?
 
1,100
0
the reduction in the power due to differentiations on [itex]x^n[/itex] are cancelled out by the x^2 in front of y"
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.

What does a linear combination of them look like?
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 ©)
 
1,100
0
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 ©)
 
Mute
Homework Helper
1,381
10
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[/tex],

which is a linear first order differential equation that you can solve using an integrating factor.
 
dextercioby
Science Advisor
Homework Helper
Insights Author
12,965
536
I'd say it's better if you simply separate variables and integrate instead of calculating the integrating factor.

Daniel.
 
1,100
0
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[/tex],

which is a linear first order differential equation that you can solve using an integrating factor.
I'd say it's better if you simply separate variables and integrate instead of calculating the integrating factor.
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 :biggrin:

The Bob (2004 )
 
HallsofIvy
Science Advisor
Homework Helper
41,738
899
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.
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.
 
Mute
Homework Helper
1,381
10
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.
What if instead of e^x you had had [itex](2+2x)w' + 6w = 0[/itex] or [itex](117 + 117x)w' + 351w = 0[/itex]? 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 [itex]x^n[/itex] in your first problem, to get [itex]e^x((1+x)w' + 3w) = 0[/itex] and note that for this equation to hold for all x, you need [itex](1+x)w' + 3w = 0[/itex].
 
Last edited:

Related Threads for: Differential Equations

Replies
5
Views
518
Replies
5
Views
795
  • Last Post
Replies
2
Views
571
  • Last Post
Replies
20
Views
2K
  • Last Post
Replies
4
Views
739
  • Last Post
Replies
2
Views
651
  • Last Post
Replies
4
Views
1K
Top