second d.e

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December 25th 2008, 08:21 PM
tsal15

second d.e

Hey all,

Season greetings :D

Merry Christmas and a Happy New Year :D

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

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 2-2e^{-2x}$

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

diff eqn: $\frac{dy}{dx} + \frac{2y}{x} = 4x$

gen eqn: $y=x^2 + \frac{c}{x^2}$ , where $p(x) = x^2$

Thank you all for your help.

tsal15
December 25th 2008, 08:45 PM
Danny

Quote:

Originally Posted by tsal15

Hey all,

Season greetings :D

Merry Christmas and a Happy New Year :D

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

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 2-2e^{-2x}$

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

diff eqn: $\frac{dy}{dx} + \frac{2y}{x} = 4x$

gen eqn: $y=x^2 + \frac{c}{x^2}$ , where $p(x) = x^2$

Thank you all for your help.

tsal15

Q1: Please explain A and B?

Q2: Yes, your answer is correct!
December 25th 2008, 09:02 PM
tsal15

Quote:

Originally Posted by danny arrigo

Q1: Please explain A and B?

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 2-2e^{-2x}$

trial function: $y = B - Ae^{-2x}$

Therefore, $-4Ae^{-2x} + 2Ae^{-2x} -2(B - Ae^{-2x}) = 2 -2e^{-2x}$

Equating Coefficients:

$-4Ae^{-2x} + 2Ae^{-2x} + 2Ae^{-2x} = -2e^{-2x}<br />$

$0A \neq -2$

Therefore, A has no solutions

Continuing to equate coefficients:

$-2B = 2$

$B = -1$

Do you follow?

Thanks Danny
December 25th 2008, 09:07 PM
Isomorphism

Quote:

Originally Posted by tsal15

Hey all,

Season greetings :D

Merry Christmas and a Happy New Year :D

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

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 2-2e^{-2x}$

I dont know what you mean by "A and B in the general solution". However I can solve the differential equation for you.

If $z = \frac{dy}{dx} - y$ , then $\frac{dz}{dx} + 2z = 2 - 2e^{-2x}$

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

I think $z = 1 - 2xe^{-2x} + C_1 e^{-2x}$ . This means $\frac{dy}{dx} - y = 1 - 2xe^{-2x} + C_1 e^{-2x}$

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

Quote:

Originally Posted by tsal15

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

diff eqn: $\frac{dy}{dx} + \frac{2y}{x} = 4x$

gen eqn: $y=x^2 + \frac{c}{x^2}$ , where $p(x) = x^2$

Thank you all for your help.

tsal15

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:

Solve: x = 1:

$x = 1 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$

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, but any kind of equation.

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

Regards,
Iso
December 25th 2008, 09:18 PM
tsal15

"If $z = \frac{dy}{dx} - y$ , then $\frac{dz}{dx} + 2z = 2 - 2e^{-2x}$ "

How is $2z = \frac{dy}{dx} - 2y$ ? I would have thought $2z = 2\frac{dy}{dx} - 2y$ , no?

"I think $z = 1 - 2xe^{-2x} + C_1 e^{-2x}$ ."

How was this achieved?

Thanks Iso :D
December 25th 2008, 09:18 PM
Danny

Quote:

Originally Posted by Isomorphism

However I can solve the differential equation for you.

If $z = \frac{dy}{dx} - y$ , then $\frac{dz}{dx} + 2z = 2 - 2e^{-2x}$

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

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

$y'' + y = \sec x?$

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
Danny

Quote:

Originally Posted by tsal15

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 2-2e^{-2x}$

trial function: $y = B - Ae^{-2x}$

Therefore, $-4Ae^{-2x} + 2Ae^{-2x} -2(B - Ae^{-2x}) = 2 -2e^{-2x}$

Equating Coefficients:

$-4Ae^{-2x} + 2Ae^{-2x} + 2Ae^{-2x} = -2e^{-2x}<br />$

$0A \neq -2$

Therefore, A has no solutions

Continuing to equate coefficients:

$-2B = 2$

$B = -1$

Do you follow?

Thanks Danny

First consider the homogenous problem

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0$

The solution is

$y = c_1 e^{x} + c_2 e^{-2x}$

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

$y_p = A + B \underbrace{x}_{bump} e^{-2x}$
December 25th 2008, 10:03 PM
tsal15

Quote:

Originally Posted by danny arrigo

First consider the homogenous problem

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0$

The solution is

$y = c_1 e^{x} + c_2 e^{-2x}$

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

$y_p = A + B \underbrace{x}_{bump} e^{-2x}$

Hey Danny,

What do you mean by 'bump'?

Also, I've lost you with this solution $y = c_1 e^{x} + c_2 e^{-2x}$ Where's this come from? :(

So, my process was wrong?

Thanks Danny
December 25th 2008, 10:12 PM
mr fantastic

Quote:

Originally Posted by tsal15

Hey Danny,

What do you mean by 'bump'?

Also, I've lost you with this solution $y = c_1 e^{x} + c_2 e^{-2x}$ Where's this come from? :(

So, my process was wrong?

Thanks Danny

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
tsal15

Quote:

Originally Posted by danny arrigo

First consider the homogenous problem

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0$

The solution is

$y = c_1 e^{x} + c_2 e^{-2x}$

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

$y_p = A + B \underbrace{x}_{bump} e^{-2x}$

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 :(

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0$

The characteristic equation: $m^2 + m - 2 = 0$

The roots to this equation: $m_1 = -2$ and $m_2 = 1$

therefore the complementary function: $x(t) = Ae^{-2t} + Be^{t}$

so my $y_p(x) = \frac{-2}{-3}xe^{-2x}$

therefore my general solution is $y(x) = \frac{2}{3}xe^{-2x} - Ae^{-2t} + Be^{t}$

Thanks again :)
December 26th 2008, 03:18 AM
mr fantastic

Quote:

Originally Posted by tsal15

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 :(

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0$

The characteristic equation: $m^2 + m - 2 = 0$

The roots to this equation: $m_1 = -2$ and $m_2 = 1$

therefore the complementary function: $x(t) = Ae^{-2t} + Be^{t}$ Mr F says: Correct in spirit I suppose. But shouldn't it be ${\color{red}y(x) = Ae^{-2x} + Be^{x}}$ ....

so my $y_p(x) = \frac{-2}{-3}xe^{-2x}$ Mr F says: This part is correct. But there's another part that you've forgotten to include. See main reply below.

therefore my general solution is $y(x) = \frac{2}{3}xe^{-2x} - Ae^{-2t} + Be^{t}$

Thanks again :)

What about the particular solution corresponding to the 2 .....

Quote:

Originally Posted by tsal15

[snip]

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = {\color{red}2} - 2e^{-2x}$

[snip]
December 26th 2008, 04:16 AM
tsal15

Quote:

Originally Posted by mr fantastic

What about the particular solution corresponding to the 2 .....

Can I just add a 2x? to the equation? if not, can you show me? I've looked through all my notes and as much as its hard to understand, i am also having difficulty locating something that will help me here...:(

$y(x) = \frac{2}{3}xe^{-2x} - Ae^{-2x} + Be^{x} + 2x$

Thank you Mr. Fantastic for your previous assistance :D
December 26th 2008, 04:38 AM
mr fantastic

Quote:

Originally Posted by tsal15

Can I just add a 2x? to the equation? if not, can you show me? I've looked through all my notes and as much as its hard to understand, i am also having difficulty locating something that will help me here...:(

$y(x) = \frac{2}{3}xe^{-2x} - Ae^{-2x} + Be^{x} + 2x$

Thank you Mr. Fantastic for your previous assistance :D

NO! Did you substitute it into the DE and check if it worked. You would see in a flash that it doesn't.

The particular solution is of the form $y = \alpha$ where $\alpha$ is a constant whose value you must find.
December 26th 2008, 06:50 AM
Danny

Quote:

Originally Posted by tsal15

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 :(

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0$

The characteristic equation: $m^2 + m - 2 = 0$

The roots to this equation: $m_1 = -2$ and $m_2 = 1$

therefore the complementary function: $x(t) = Ae^{-2t} + Be^{t}$

so my $y_p(x) = \frac{-2}{-3}xe^{-2x}$

therefore my general solution is $y(x) = \frac{2}{3}xe^{-2x} - Ae^{-2t} + Be^{t}$

Thanks again :)

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

$y'' + y' -2y = 2 - 2e^{-2x}$

it has powers $x^0$ and exponential's $e^{-2x}.$ Therefore one would try a guess of

$y = A + B e^{-2x}$

Taking derivatives

$y' = -2 B e^{-2x},\;\;\;y'' = 4 B e^{-2x},$

and substituting gives

$y'' + y' - 2y = 4 B e^{-2x} + \left( -2 B e^{-2x}\right) - 2 \left( A + B e^{-2x}\right) = 2 - 2e^{-2x}$

Expanding gives $-2A = 2 - 2e^{-2x}$ (yes, the B's cancel but why?) The reason is that $y = Be^{-2x}$ is part of the complimentary solution so

$y'' + y' - 2y = 0$ identically! So in this case, what do we do for a guess. Let us onsider the following first order example

$(1)\;\;\; y' - y = e^x$

The complimentary equation is

$y' - y = 0$

the characteristic equations is

$m - 1 = 0$

the complimentary solution is

$y = e^x$

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

$y = c e^x + x\,e^x$

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

$y = A + Bx e^{-2x}$

$y' = B e^{-2x} -2x Bx e^{-2x}$
$y'' = -4B e^{-2x} + 4x Bx e^{-2x}$

substituting gives

$y'' + y' - 2y = -4B e^{-2x} + 4 Bx e^{-2x} + \left( B e^{-2x} -2 Bx e^{-2x} \right) - 2 \left( A + Bx e^{-2x} \right)=$
$= 2 - 2e^{-2x}$

and when we expand we get

$-3B e^{-2x} - 2 A = 2 - 2e^{-2x}$

from which we obtain $A = -1,\;\;B = \frac{2}{3}$ and the particular solution

$y_p = - 1 +\frac{2}{3} \,e^{-2x}$
December 26th 2008, 05:19 PM
Prove It

Quote:

Originally Posted by tsal15

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 :(

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0$

The characteristic equation: $m^2 + m - 2 = 0$

The roots to this equation: $m_1 = -2$ and $m_2 = 1$

therefore the complementary function: $x(t) = Ae^{-2t} + Be^{t}$

so my $y_p(x) = \frac{-2}{-3}xe^{-2x}$

therefore my general solution is $y(x) = \frac{2}{3}xe^{-2x} - Ae^{-2t} + Be^{t}$

Thanks again :)

Where did the t's come from?

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