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- December 25th 2008, 08:21 PM #1

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## second d.e

Hey all,

Season greetings

Merry Christmas and a Happy New Year

Does the following differential equation not have a solution for A, and is B = -1 in the general solution ?

Also does the general solution of the following differential equation look anything like what I've given?

diff eqn:

gen eqn: , where

Thank you all for your help.

tsal15

- December 25th 2008, 08:45 PM #2

- December 25th 2008, 09:02 PM #3

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- December 25th 2008, 09:07 PM #4
I dont know what you mean by "A and B in the general solution". However I can solve the differential equation for you.

If , then

The above equation is easily solvable using the "integrating factor" method.

I think . This means

The above equation is easily solvable, again, using the "integrating factor" method.

The answer is right.

**A general tip:**

*While checking, if any obtained answer to an equation is the solution to the original question, you can always help yourself by substituting the answer back and verifying whether the answer satisfies the equation.*

Moreover one should always do this(subbing back) while solving equations, since extraneous roots could have crept in at some steps.

Here's a silly example that makes its point:

: x = 1:**Solve**

But observe that x = -1 doesnt satisfy x = 1. This happened because the second step is not reversible

Thus even in exams you can sub back and check, and be confident about your answer. Note that this tip is not just for differential equations, butkind of equation.__any__

Why depend on others when you can help yourself, right?

Regards,

Iso

- December 25th 2008, 09:18 PM #5

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- December 25th 2008, 09:18 PM #6
Hey Iso, most people can't simply introduced a new variable (as you have) as to reduce the second order problem to one that is first order. For example, what new variable would you introduce for the problem

The idea of seeking a particular solution and reducing the problem to one that is homogeneous I think is easier for most.

- December 25th 2008, 09:30 PM #7
First consider the homogenous problem

The solution is

Next question, is the complimentary solution a part of the nonhomogeneous term. In this case yes! To use the method of undetermined coefficients it is necessary to "bump up" the solution and seek a particluar solution of the form

- December 25th 2008, 10:03 PM #8

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- December 25th 2008, 10:12 PM #9
You're expected to know the stuff discussed in post #3 of this thread: http://www.mathhelpforum.com/math-he...-equation.html

And you're expected to know how to construct a particular solution. Have you reviewed your class notes and textbook?

- December 26th 2008, 02:51 AM #10

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Hey, i got a few questions sorry

1 - I haven't come across the term "Bump Up" in all my notes and texts, can you explain?

2 -WELL I've tried to do it again, here's what I've got. PLEASE tell me where I've gone WRONG

The characteristic equation:

The roots to this equation: and

therefore the complementary function:

so my

therefore my general solution is

Thanks again

- December 26th 2008, 03:18 AM #11

- December 26th 2008, 04:16 AM #12

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- December 26th 2008, 04:38 AM #13

- December 26th 2008, 06:50 AM #14
Yes, your complimentary solution is right (via your characteristic equation) except you want x's and not t's. Let me explain a little bit more about the construction of this particular solution. There are primary two methods to do this (1) the method of undetermined coefficients and (2) the variation of parameters. The first method is usually easier to use but involves some educated guessing. The method basic seeks the form of the particular solution by looking at the rhs of the ODE. If it has power's guess power's, exponential's guess exponential's, sin's and cos's guess sin's and cos's. In your example

it has powers and exponential's Therefore one would try a guess of

Taking derivatives

and substituting gives

Expanding gives (yes, the B's cancel but why?) The reason is that is part of the complimentary solution so

identically! So in this case, what do we do for a guess. Let us onsider the following first order example

The complimentary equation is

the characteristic equations is

the complimentary solution is

However, since eqn. (1) is linear we solve giving

In this solution we have two parts, the first the complimentary and the second the particular. We notice the extra x so in order to guess a particular solution when rhs contains the complimentary solution we "bump" up the exponential in our first guess and use

substituting gives

and when we expand we get

from which we obtain and the particular solution

- December 26th 2008, 05:19 PM #15