Reduction of Order Problem

[praecox]

Hey, guys. I'm having a hard time with a 2nd ODE reduction of order problem.

The DE is
t²y'' - 4ty' + 6y = 0, with y₁ = t².

So I set up my y = t²v(t); y' = t²v' + 2tv; y'' = t²v'' + 4tv' + 2v.
I then substituted back in and got:
t⁴v'' = 0

This is where I'm getting stuck. The book says that the answer I'm looking for is y₂ = t³, but I can't seem to get there from here.

I don't think I can use an integrating factor here (if I'm wrong there let me know), so I tried separating the variables, but that got me no where. I think I did it wrong though.

Here was my separation of variables attempt:
letting w = v', so w' = v'':
t⁴dw/dt = 0
dw = t^-4 dt
Integrating both sides (and taking constants of integration to be zero):
w = -1/3 t^-3
since w = v':
dv = -1/3 t^-3 dt
Integrating again to find v and still no where near t³.

Any ideas? Anybody? :/

 

[Some Pig]

Let y=tⁿ
n(n-1)tⁿ-4ntⁿ+6tⁿ=0
(n-2)(n-3)tⁿ=0
n=2 or 3

 

[matematikawan]

Hey, guys. I'm having a hard time with a 2nd ODE reduction of order problem.

The DE is
t²y'' - 4ty' + 6y = 0, with y₁ = t².

So I set up my y = t²v(t); y' = t²v' + 2tv; y'' = t²v'' + 4tv' + 2v.
I then substituted back in and got:
t⁴v'' = 0


Any ideas? Anybody? :/

v"(t)=0 hence v(t) is linear. Choose v(t)=t so that y = t²v(t)=t³.

 

[praecox]

That makes so much sense now! Thank you so much! :D

 

[blue_raver22]

Hello there,

I understand your frustration with this reduction of order problem. It can definitely be a tricky concept to grasp at first. Let's take a closer look at your approach and see if we can figure out where things went wrong.

First, you correctly set up the substitution of y = t^2v(t). This is a common method for solving second order differential equations. However, when you substituted back into the original equation, you made a mistake. The correct substitution should be:

t^4v'' - 4t^3v' + 6t^2v = 0

You missed the t^3 term in the second term. This small mistake is what caused you to get stuck later on.

Now, let's continue with your separation of variables attempt. You correctly substituted w = v' and then integrated both sides. However, when you integrated both sides, you forgot to include the t term in the integration of the left side. It should be:

∫ t^4dw = ∫ t^-3dt

This will give you:

t^4w = -1/2 t^-2 + C

Now, when you solve for w, you should get:

w = -1/2 t^-6 + C

Remember, w = v', so we need to integrate one more time to find v. This gives us:

v = -1/10 t^-5 + Ct + D

Finally, we can substitute this back into our original substitution:

y = t^2v = t^2(-1/10 t^-5 + Ct + D) = -1/10 t^-3 + Ct^3 + Dt^2

This is the general solution to the differential equation. Now, we can use the initial condition y1 = t^2 to find the specific solution. Plugging in t^2 for y and solving for C and D, we get:

C = 0 and D = 1

Therefore, the specific solution to the differential equation is:

y = t^3

I hope this helps you understand the reduction of order method better. Keep practicing and don't get discouraged. As a scientist, perseverance is key in solving problems and finding solutions. Good luck!