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

[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' = nx^n-1 and y'' = n(n-1)x^n-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 ©)

 

[cristo]

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]

Show us what you get after making the substitutuons for

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

 

[The Bob]

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

$$y = x^n$$ 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 ©)

 

[morson]

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

 

[Mute]

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 $ax^2y'' + by' + cy = 0$, 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 [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, $y = Ax^3 + Bx^4$.

So it appears you knew what you were doing, you just weren't sure of it.

 

[HallsofIvy]

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?

 

[The Bob]

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.

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 ©)

 

[The Bob]

Whilst I am on a role, I have another question before tomorrow.

Given that y = e^x 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 = ue^x, differentiate this and substitued in. Then I have let w = u' and got down to:

w'(e^x + xe^x) + 3e^xw = 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]

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

$$(1+x)w' + 3w = 0$$,

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

 

[dextercioby]

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

Daniel.

 

[The Bob]

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

$$(1+x)w' + 3w = 0$$,

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 e^x 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 )

 

[HallsofIvy]

Whilst I am on a role, I have another question before tomorrow.

Given that y = e^x 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 e^x 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]

May I thank you both. Although I do not understand completely why you can simply divide through by e^x 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 $(2+2x)w' + 6w = 0$ 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 $(1+x)w' + 3w = 0$.