2nd order differential equation using reduction of order

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Homework Help Overview

The discussion revolves around finding a second solution to a second-order differential equation using the method of reduction of order. The specific equation is t²y'' - 4ty' + 6y, with the given solution y1(t) = t² for t > 0.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the substitution y = vt² and derive expressions for y', y'', leading to the equation t⁴v'' = 0. There is uncertainty about the next steps, particularly regarding dividing by y and integrating. Some participants question whether v'' = 0 implies a straightforward integration process.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the derived equation and attempting to clarify the integration process. Some guidance has been offered regarding the integration of the second-order differential equation, but there is no consensus on the final form of the solution.

Contextual Notes

Participants express confusion about the integration steps and the implications of the boundary conditions, particularly regarding the positivity of t. There is a repeated emphasis on the need to integrate twice with respect to t, but the exact approach remains unclear to some.

efghi
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Homework Statement



Use the method of reduction of order to find a second solution of the given differential equation

t2y'' - 4ty' + 6y, t>0; y1(t) = t2


The Attempt at a Solution



Here's what I have so far:
y = vt2
y' = 2tv + t2v'
y'' = 2v + 4tv' + t2v''

so

t2 (2v + 4tv' + t2v'') - 4t (2tv + t2v') + 6(vt2) = 0

t4v'' = 0

I'm stuck here. Am I supposed to divide by y, which would make it t2v'' = 0
wouldn't that make v'' = 0 anyways? What do I do now? I know I'm supposed to integrate it some how but I am just utterly lost from here...

PLEASE HELP!
 
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efghi said:

Homework Statement



Use the method of reduction of order to find a second solution of the given differential equation

t2y'' - 4ty' + 6y, t>0; y1(t) = t2


The Attempt at a Solution



Here's what I have so far:
y = vt2
y' = 2tv + t2v'
y'' = 2v + 4tv' + t2v''

so

t2 (2v + 4tv' + t2v'') - 4t (2tv + t2v') + 6(vt2) = 0

t4v'' = 0

I'm stuck here. Am I supposed to divide by y, which would make it t2v'' = 0
wouldn't that make v'' = 0 anyways? What do I do now? I know I'm supposed to integrate it some how but I am just utterly lost from here...
You're absolutely spot on here. Let's look at your final line

t4v''=0

Now the boundary conditions state that t is positive, hence this means that t4 must be non-zero. Furthermore, this implies that

v''=0

Now you have a second order differential equation

[tex]\frac{d^2 v}{dt^2}=0[/tex]

Simply integrate twice with respect to t. Do you follow?
 
So the 2nd general solution should be y2 = t6 / 18 ?
 
efghi said:
So the 2nd general solution should be y2 = t6 / 18 ?
How did you get that?
 
integrate twice with respect to t

integral of t4v'' = t5 / 3
integral of t5 / 3 = t6 / 18?
 
efghi said:
integrate twice with respect to t

integral of t4v'' = t5 / 3
integral of t5 / 3 = t6 / 18?
That's wrong. As I said in my previous post, you need only solve the ODE

v''=0

And then multiply the solution by y1 to obtain the general solution.
 
How do i solve the ODE?
 
efghi said:
How do i solve the ODE?
As I said before integrate twice with respect to t.
 

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