# Second Order Differential Equation

• Nov 16th 2008, 01:08 AM
panda*
Second Order Differential Equation
Hi!

How do I determine the particular integral for second order differential equations with mixed f(x)s? Like, it's neither strictly polynomial, trigonometry or exponential.

Example of questions,

(i) $y'' - 4y' + 5y = (16x + 4)e^{3x}$
(ii) $y'' + 3y' = (10x + 6)sin x$
(iii) $y'' - 2y' + 4y = 541e^{2x}cos5x$

Thank you! (:
• Nov 16th 2008, 08:14 AM
shawsend
Hey, each of those have right members which are particular solutions to some homogeneous differential equation. For example, the first one has a right member which is a solution to the equation:

$(D-3)^2 y=0$

So the operator $(D-3)^2$ becomes an "annihilation" operator that we can apply to both side of the equation to convert it to a homogeneous equation:

$(D-3)^2 (D^2-4D+5)y=0$

This is the method of undetermined coefficients. Are you familiar with that method? Try it first on some simple ones. Any DE book should have a section on this subject.
• Nov 16th 2008, 08:49 PM
panda*
Sorry but I do not understand what are you talking about at all. :(
• Nov 16th 2008, 08:56 PM
Chris L T521
Quote:

Originally Posted by shawsend
Hey, each of those have right members which are particular solutions to some homogeneous differential equation. For example, the first one has a right member which is a solution to the equation:

$(D-3)^2 y=0$

So the operator $(D-3)^2$ becomes an "annihilation" operator that we can apply to both side of the equation to convert it to a homogeneous equation:

$(D-3)^2 (D^2-4D+5)y=0$

This is the method of undetermined coefficients. Are you familiar with that method? Try it first on some simple ones. Any DE book should have a section on this subject.

Aha!! I'm not the only one to know about the Annihilator approach :D

Quote:

Originally Posted by panda*
Sorry but I do not understand what are you talking about at all. :(

Read post #6 and #7 here to see how to tackle equations like these.

--Chris
• Nov 16th 2008, 09:05 PM
Math_Helper
1. First you must solve the homogeneous equations
(i) y''-4y-+5y=0
(ii) y''+3y'=0
(iii) y''-2y'+4y=0
and you will find the general solutions for them.
2. For non homogeneous equations you have to find the partial solutions according to what is on the right side of the equations
(i) http://www.mathhelpforum.com/math-he...9a1005f4-1.gif
(ii) http://www.mathhelpforum.com/math-he...abc676a5-1.gif
(iii) http://www.mathhelpforum.com/math-he...fa9942ab-1.gif
3. The final solution of your equations is the sum of the general solutions from 1st point and partial solutions from the 2nd point.
• Nov 17th 2008, 12:36 AM
panda*
Thank you for referrals of posts and helps.

But I still don't really get it after I read the posts :(

Generally, I do know how to handle second order differential equations if f(x) was one of the normal terms like, purely exponential, or trig or polynomials. But when it is multiplied together like that, I can't immediately identify the pattern of the particular integral.

For example, the particular integral form for an exponential function of $f(x) = e^{kx}$ would be $y_p = pe^{2x}$ and we differentiate and sub it back in to compare coefficients in order to obtain p.

How do I deal when f(x) is not purely of those forms but multiplied together?

Thank you.
• Nov 17th 2008, 02:53 AM
shawsend
I'll work the first one using undetermined coefficients; differential equations open a unique window into the universe such that all her secrets are revealed:

$(D-3)^2(D^2-4D+5)y=0$

The general solution from the auxiliary equation is then:

$y=c_1 e^{3x}+c_2 e^{3x}+c_4 e^{2x}\cos(x)+c_5 e^{2x}\sin(x)$

with the desired solution (I'm taking this right out of Rainville almost word for word):

$y=y_c+y_p$

where $y_c=c_4 e^{2x}\cos(x)+c_5 e^{2x}\sin(x)$. Then there must be a particular solution of the original equation containing at most the remaining terms: $y_p=Ae^{3x}+Bxe^{3x}$. That's the undetermined coefficients which can be determined by substituting this $y_p$ into the original DE:

When I do that I get:

$2Ae^{3x}+2Bxe^{3x}+2Be^{3x}=16xe^{3x}+4e^{3x}$

Equating coefficients, I get $B=8$ and $A=-6$. Then the general solution of $y''-4y'+5y=(16x+4)e^{3x}$ is:

$y(x)=-6e^{3x}+8xe^{3x}+c_1e^{2x}\cos(x)+c_2e^{2x}\sin(x)$
• Nov 17th 2008, 06:57 PM
panda*
Although I still do not understand the method you recommended, but I'm still thankful for your generous help and time! I figured out a different approach to do it already :) But still thanks anyway!
• Nov 18th 2008, 04:23 AM
shawsend
Quote:

Originally Posted by panda*
I figured out a different approach to do it already

Close enough. :)