How to Solve Second Order Differential Equations with Non-Constant Coefficients?

The Bob
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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 ©)
 
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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!
 
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Show us what you get after making the substitutuons for

y = x^n and its derivatives.
 
cristo said:
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!

Integral said:
Show us what you get after making the substitutuons for

y = x^n 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 ©)
 
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Wouldn't you just leave it as "y = x^4, y = x^3"?
 
The Bob said:
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 ax^2y&#039;&#039; + by&#039; + cy = 0, the reduction in the power due to differentiations on x^n are canceled 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&#039;s why n = 3, 4.<br /> <br /> 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.<br /> <br /> So it appears you knew what you were doing, you just weren&#039;t sure of it.
 
The Bob said:
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?
 
Mute said:
the reduction in the power due to differentiations on x^n are canceled 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.

HallsofIvy said:
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 ©)
 
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
Keep in mind that e^x is never zero, so you can divide it out of your equation in w to get

(1+x)w&#039; + 3w = 0,

which is a linear first order differential equation that you can solve using an integrating factor.
 
  • #11
I'd say it's better if you simply separate variables and integrate instead of calculating the integrating factor.

Daniel.
 
  • #12
Mute said:
Keep in mind that e^x is never zero, so you can divide it out of your equation in w to get

(1+x)w&#039; + 3w = 0,

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

dextercioby said:
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 )
 
  • #13
The Bob said:
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.
 
  • #14
The Bob said:
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 (2+2x)w&#039; + 6w = 0 or (117 + 117x)w&#039; + 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&#039; + 3w) = 0 and note that for this equation to hold for all x, you need (1+x)w&#039; + 3w = 0.
 
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