2nd order differential equation using reduction of order

In summary: Do you follow?In summary, the conversation discusses using the method of reduction of order to find a second solution of a given differential equation. The attempt at a solution involves finding the general solution by integrating twice with respect to t. The final solution is found to be y2 = t6 / 18.
  • #1
efghi
4
0

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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  • #2
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?
 
  • #3
So the 2nd general solution should be y2 = t6 / 18 ?
 
  • #4
efghi said:
So the 2nd general solution should be y2 = t6 / 18 ?
How did you get that?
 
  • #5
integrate twice with respect to t

integral of t4v'' = t5 / 3
integral of t5 / 3 = t6 / 18?
 
  • #6
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.
 
  • #7
How do i solve the ODE?
 
  • #8
efghi said:
How do i solve the ODE?
As I said before integrate twice with respect to t.
 

What is a 2nd order differential equation?

A 2nd order differential equation is a mathematical equation that involves the second derivative of an unknown function. It is commonly used in physics and engineering to model systems with acceleration and other complex behaviors.

What is reduction of order?

Reduction of order is a technique used to solve 2nd order differential equations that involves substituting a new variable for the original unknown function. This reduces the order of the differential equation to a 1st order equation, which can be solved using standard techniques.

How do you solve a 2nd order differential equation using reduction of order?

To solve a 2nd order differential equation using reduction of order, you first identify a known solution to the equation, called the “particular solution”. Then, you use this solution to substitute a new variable in place of the original unknown function. Finally, you solve the resulting 1st order equation to find the general solution.

What are some common applications of 2nd order differential equations?

2nd order differential equations are commonly used in physics and engineering to model systems with acceleration, such as pendulums, springs, and electrical circuits. They are also used in finance and economics to model growth and decay processes.

What are some techniques for solving 2nd order differential equations?

In addition to reduction of order, other techniques for solving 2nd order differential equations include separation of variables, variation of parameters, and the method of undetermined coefficients. Each technique is suited for different types of differential equations and can provide different insights into the behavior of the system being modeled.

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