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 January 21st, 2015, 09:04 PM #1 Newbie   Joined: Jan 2015 From: world Posts: 11 Thanks: 0 differential equations and matrix question 1) y''+2y'+5y=sin2x a) what is the general solution ? b) y(0)=1 , y'(0)=0 solution ? 2) Find the eigen frequency?
 January 22nd, 2015, 02:12 AM #2 Newbie   Joined: Jan 2015 From: world Posts: 11 Thanks: 0 Does not solve ?
 January 22nd, 2015, 02:56 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra For the first, you will start by using the characteristic (quadratic) polynomial to find the complementary solution. Then you can find a particular solution by forming a trial solution with unknown coefficients which you can solve for. The sum of the particular and complementary solutions is the general solution. This contains two arbitrary constants. In the second part you use the given initial conditions to find the value of those constants.
 January 22nd, 2015, 03:08 AM #4 Newbie   Joined: Jan 2015 From: world Posts: 11 Thanks: 0 Can you make the solution?
 January 22nd, 2015, 03:43 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Yes, thank you. Can you?
 January 22nd, 2015, 08:53 PM #6 Newbie   Joined: Jan 2015 From: world Posts: 11 Thanks: 0 2 question ?
 January 23rd, 2015, 03:54 AM #7 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 What is your purpose is asking this question? If you are taking, or have taken, a course in elementary differential equations, you should be able to find the general solution yourself, especially with V8archie's suggestion. If not, I don't see how being given the solution will help you. The equation, y''+2y'+5y=sin2x, is a "linear, non- homogeneous second order differential equation with constant coefficients". There is a pretty standard method of solution: First solve the "associated homogeneous equation"y''+ 2y'+ 5y= 0. Its "characteristic equation" is $\displaystyle r^2+ 2r+ 5= 0$. Can you solve that equation? How do its roots help you solve the differential equation? Now, you need to find a single solution to the entire equation. Since this is a an equation with constant coefficients and the "non-homogeneous part" is sin(2x), I recommend trying a solution of the form "y= A cos(2x)+ B sin(2x)". Put that into the equation and see what A and B must be to satisfy that equation. For the second problem, let x be the distance from the left wall to the 2 kg mass and let y be the distance from the left wall to the 4 kg mass. You don't give a width for either mass or a "rest length" for any of the springs so I will assume those are 0. Use "force= mass times acceleration". The acceleration of the 2 kg mass is $\displaystyle \frac{d^2x}{dt^2}$ so "mass times acceleration" is $\displaystyle 2\frac{d^2x}{dt^2}$. Now for the force. The net force on the 2 kg mass is the difference between the two spring forces and "spring force" is the "spring constant times the amount of stretch" or, because I am taking the "rest length" to be 0, just 'spring constant times the length". The leftmost spring goes from the left wall to the 2kg mass so its length is 0. The force is 1 times x. The middle spring goes from the 2kg mass to the 4 kg mass so its length is y- x. The force is 2(y- x)= 2y- 2x. Since the first force is directed to the left and the second to the right, the net force is, as I said, the difference between those two: 2y- 2x- x= 2y- x. That is, $\displaystyle 2\frac{d^2x}{dt^2}= 2y- x$. Do the same analysis on the 4 kg mass to get the other equation. I presume you know what an "eigen frequency" is. (If you do, you should know there is more than one and you are really asking for "eigen frequencies".) Last edited by Country Boy; January 23rd, 2015 at 04:17 AM.
 January 23rd, 2015, 06:08 AM #8 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions Yakamoz29, you should be aware that many many many people come onto this forum and simply expect solutions to their maths problems, usually to get answers for some homework or a graded assignment. We are not a homework service, so most of us will not just state the answers for you. However, if you find something difficult, most of us are happy to help you through it. Have a go at the problem yourself and let us know specifically what steps you find difficult/tricky. If you don't know where to start, we can provide links to online resources that run through the mathematical methods you need to know or, in some instances, even explain everything ourselves. We all know how hard maths can be, so we don't care if you make a mistake or do something totally wrong. In short; make an effort and we will guide you in the right direction.
 January 24th, 2015, 01:59 AM #9 Newbie   Joined: Jan 2015 From: world Posts: 11 Thanks: 0 eigen frequencies. Can you question 2 solution? I could not solve.
 January 24th, 2015, 03:25 AM #10 Newbie   Joined: Jan 2015 From: world Posts: 11 Thanks: 0 how to contine ?

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