2nd order DEQ: weird solution method

In summary, the student attempted to find a solution to the homework statement y'' + p(t)y' + q(t)y = 0 by using the method of variation of parameters. However, they were not able to find a solution and were instead able to find a solution to the equation y(t) = e^(2t).
  • #1
fusi0n
16
0

Homework Statement



Suppose that u(t) is a solution to

y'' + p(t)y' + q(t)y = 0

Suppose a second solution has the form

y(t) = m(t)u(t)

where m(t) is an unknown function of t. Derive a first order linear differential equation for m'(t).

Suppose y(t) = e^(2t). Use the method above to find a second solution to this equation.

The Attempt at a Solution



I tried:

y = m*u
=> m = y/u
=> m' = (y'/u) - (y/(u^2))*u'

There isn't anything in my book or in my class notes that describes the technique to be used here. Does anyone have any suggestions on how to get started?
 
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  • #3
Surely this is recommended where e^{2t} is a solution to the original 2nd order diff. eq. Right? Then, like Mindscrape says, it's variation of parameters.
 
  • #4
try using this equation:

Y2=y1*integral{[e^(-integral p(t))dt]}/[(y1)^2]dt

y1=y(t)=e^(2t)

then the general solution is: y=c1y1+c2y2
 
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  • #5
And also to add, your method works best when the p(t)y' value is excluded from the equation. You can solve your problem by the method you mentioned above, but it cost much time and leaves much room for mistakes, and you are essentially guessing by trial and error by multiplying u(t) by respective powers of x, or m, as you would put it. So therefore, for the sake of your precious time and of doing unneeded amount of work, you should try this equation. Tell me how it works, I believe you will learn to like it. And, later you will learn to solve these problems without the guesswork, but the problems get very tedious, believe me... I'm not a big fan of solving these equations by hand.
 
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  • #6
this is an inhomogeneous differential equation. the method of variation of parameters is for homogeneous. in the wikipedia article it is said to "determine the homogeneous solution using a method of your choice"

as for as the other method is concerned: how can I use this method to derive a general first order linear differential equaiton for m'(t)? by the way mathgician, this method works great for finding the second solution!
 
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  • #7
never mind, mathgician, i out the way to find m'(t). your input helped a lot. i will post my solution to the problem if anyone is interested.
 

What is a 2nd order DEQ?

A 2nd order DEQ stands for a second-order differential equation. It is an equation that relates a function to its derivatives. It is typically in the form of y'' = f(x, y, y'), where y' is the first derivative of y and y'' is the second derivative of y.

What makes the solution method for 2nd order DEQs weird?

The solution method for 2nd order DEQs can be considered weird because it involves finding two separate solutions to the equation and combining them to form the general solution. This is in contrast to first-order DEQs where only one solution is needed.

Why is it important to understand 2nd order DEQs?

Understanding 2nd order DEQs is important because they are used to model many real-world phenomena in various fields such as physics, engineering, and economics. Being able to solve these equations allows us to make predictions and understand the behavior of systems.

What are some common methods for solving 2nd order DEQs?

Some common methods for solving 2nd order DEQs include the method of undetermined coefficients, variation of parameters, and Laplace transform. Each method has its own advantages and is suitable for different types of equations.

Are there any applications of 2nd order DEQs in everyday life?

Yes, 2nd order DEQs have many applications in everyday life. They are used to model the motion of objects, such as a swinging pendulum or a bouncing ball. They are also used in electrical circuits to analyze the flow of current. Understanding these equations can help us design and optimize various systems in our daily lives.

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