What is calculus and differential equations and how are they used?
Q. I graduated with a degree in math. I took four semesters of calculus, plus differential equations. I can do the math with A's, but I still have no idea exactly what calculus and differential equations ARE and why you would use them.
Asked by bamakathy - Mon Jan 14 11:36:08 2008 - - 2 Answers - 1 Comments
A. Calculus is the mathematics of infinitesimal change. Since you're familiar with calculus, for example, "dy/dx" means the instantaneous rate of change in y with respect to the instantaneous rate of change in x. Integrals, or "anti-derivatives" are sums of these infinitesimal parts. With them, you can find the areas under different curves and such. Differential equations are very useful for when you know quantities that are changing, or you're looking to solve something that's "in flux." I hope this helps.
Answered by John - Mon Jan 14 11:42:51 2008
Q. I graduated with a degree in math. I took four semesters of calculus, plus differential equations. I can do the math with A's, but I still have no idea exactly what calculus and differential equations ARE and why you would use them.
Asked by bamakathy - Mon Jan 14 11:36:08 2008 - - 2 Answers - 1 Comments
A. Calculus is the mathematics of infinitesimal change. Since you're familiar with calculus, for example, "dy/dx" means the instantaneous rate of change in y with respect to the instantaneous rate of change in x. Integrals, or "anti-derivatives" are sums of these infinitesimal parts. With them, you can find the areas under different curves and such. Differential equations are very useful for when you know quantities that are changing, or you're looking to solve something that's "in flux." I hope this helps.
Answered by John - Mon Jan 14 11:42:51 2008
How many people in the US have taken differential equations?
Q. What is the percentage of the US population that have made it all the way to Differential Equations?
Asked by Abdul Matin - Mon Mar 2 23:43:33 2009 - - 2 Answers - 0 Comments
A. Needless to say, this is something you can only estimate. Differential equations is taught at university level. So we'll start with percentage that has attended university: Wikipedia says: The 2006 American Community Survey conducted by the United States Census Bureau found that 19.5 percent of the population had attended college but had no degree, 7.4 percent held an associate's degree, 17.1 percent held a bachelor's degree, and 9.9 percent held a graduate or professional degree. Let's just take the 27 % with bachelor's or advanced degrees, since differential equations is at least a second year course. Then we have to determine how many are either math, science, or engineering students. Someone who calls himself "Captain Capitalism"… [cont.]
Answered by MathMan TG - Tue Mar 3 01:49:39 2009
Q. What is the percentage of the US population that have made it all the way to Differential Equations?
Asked by Abdul Matin - Mon Mar 2 23:43:33 2009 - - 2 Answers - 0 Comments
A. Needless to say, this is something you can only estimate. Differential equations is taught at university level. So we'll start with percentage that has attended university: Wikipedia says: The 2006 American Community Survey conducted by the United States Census Bureau found that 19.5 percent of the population had attended college but had no degree, 7.4 percent held an associate's degree, 17.1 percent held a bachelor's degree, and 9.9 percent held a graduate or professional degree. Let's just take the 27 % with bachelor's or advanced degrees, since differential equations is at least a second year course. Then we have to determine how many are either math, science, or engineering students. Someone who calls himself "Captain Capitalism"… [cont.]
Answered by MathMan TG - Tue Mar 3 01:49:39 2009
What is the difference between Linear Algebra and Differential Equations?
Q. i'm taking a course at the Harvard summer school for credits at another institution however at that instituition I only need credits for Differential Equations and Matricies which is what the course is called. However will this be the same thing as Linear Algebra and Differential Equations? i'm not familiar with the terms because this will be an introduction course for me as I only just completed multivarible calculus or (calculus 3).
Asked by Cougar J - Wed May 7 03:14:57 2008 - - 5 Answers - 0 Comments
A. They seem equivalent. The linear algebra that you are taught in Differential equations, is just matrix theory. You are only taught the linear algebra you need, so you can apply it to solving differential equations. To be more specific, you will be needing to solve system of differential equations, which you will use matrix theory to solve.
Answered by NBL - Wed May 7 03:24:36 2008
Q. i'm taking a course at the Harvard summer school for credits at another institution however at that instituition I only need credits for Differential Equations and Matricies which is what the course is called. However will this be the same thing as Linear Algebra and Differential Equations? i'm not familiar with the terms because this will be an introduction course for me as I only just completed multivarible calculus or (calculus 3).
Asked by Cougar J - Wed May 7 03:14:57 2008 - - 5 Answers - 0 Comments
A. They seem equivalent. The linear algebra that you are taught in Differential equations, is just matrix theory. You are only taught the linear algebra you need, so you can apply it to solving differential equations. To be more specific, you will be needing to solve system of differential equations, which you will use matrix theory to solve.
Answered by NBL - Wed May 7 03:24:36 2008
Express a second order differential equation as a coupled pair of first-order differential equations?
Q. Consider the second order differential equation: d^2y/dx^2 + x(dy/dx) + (1+x^2)y = 0, subject to y(1) = 1 and y'(1) = 0 Express this equation as a couple pair of first-order differential equations by letting u(x) = y(x) and v(x) = y'(x) and state the corresponding initial conditions. An answer for this would be much appreciated.
Asked by Jason W - Wed Dec 12 20:10:46 2007 - - 1 Answers - 0 Comments
A. This would be u ' = 0*u + v v ' = (1+x^2) * u + x * v Imagine vector/matrix notation (which is difficult to write here)
Answered by lienad14 - Wed Dec 12 20:31:09 2007
Q. Consider the second order differential equation: d^2y/dx^2 + x(dy/dx) + (1+x^2)y = 0, subject to y(1) = 1 and y'(1) = 0 Express this equation as a couple pair of first-order differential equations by letting u(x) = y(x) and v(x) = y'(x) and state the corresponding initial conditions. An answer for this would be much appreciated.
Asked by Jason W - Wed Dec 12 20:10:46 2007 - - 1 Answers - 0 Comments
A. This would be u ' = 0*u + v v ' = (1+x^2) * u + x * v Imagine vector/matrix notation (which is difficult to write here)
Answered by lienad14 - Wed Dec 12 20:31:09 2007
Can someone give me a description of how to solve coupled differential equations?
Q. Could someone describe how to solve the following set of differential equations? (1) dx/dt = -x(t) - y(t) (2) dy/dt = x(t) - y(t) Not too complicated, I need to know how to solve this type of problem in general more than just these ones here. Thanks.
Asked by Zyzyx - Wed Apr 1 22:37:14 2009 - - 2 Answers - 0 Comments
A. Michael M's answer is correct in that using matrix exponential methods is the standard way these problems are solved. If you have not been introduced to such concepts yet, you can still solve them by converting the two coupled first-order equations into a single second-order equation involving just one of the dependent variables, and solving for that. To see how this works, we have: dy/dt = x(t) - y(t) Take the derivative of this with respect to t: y'' = x' - y' From the first of the original equations, we know that x' = -x - y, so: y'' = -x - y - y' This still involves x, but from the second of the original equations, we know that: x = y' + y so y'' = - (y' + y) - y - y' y'' = -2y' - 2y y'' + 2y' + 2y = 0 This is now a… [cont.]
Answered by hfshaw - Fri Apr 3 18:59:43 2009
Q. Could someone describe how to solve the following set of differential equations? (1) dx/dt = -x(t) - y(t) (2) dy/dt = x(t) - y(t) Not too complicated, I need to know how to solve this type of problem in general more than just these ones here. Thanks.
Asked by Zyzyx - Wed Apr 1 22:37:14 2009 - - 2 Answers - 0 Comments
A. Michael M's answer is correct in that using matrix exponential methods is the standard way these problems are solved. If you have not been introduced to such concepts yet, you can still solve them by converting the two coupled first-order equations into a single second-order equation involving just one of the dependent variables, and solving for that. To see how this works, we have: dy/dt = x(t) - y(t) Take the derivative of this with respect to t: y'' = x' - y' From the first of the original equations, we know that x' = -x - y, so: y'' = -x - y - y' This still involves x, but from the second of the original equations, we know that: x = y' + y so y'' = - (y' + y) - y - y' y'' = -2y' - 2y y'' + 2y' + 2y = 0 This is now a… [cont.]
Answered by hfshaw - Fri Apr 3 18:59:43 2009
Are differential equations really precise in measuring air resistance?
Q. Just wondering...Everytime our instructor has to stop and says "ignore air resistance that will require differential equations which you will not get into until your engineering and physics classes and besides its complicated stuff." What are differential equations really? I always see two variables piled up next to each other and one side equals to another side plus a constant. Differential measure the rate of change, I know that, but how precise is it if there's just a constant added to it each time
Asked by crazy girl - Wed Mar 11 03:45:27 2009 - - 2 Answers - 0 Comments
A. Only if there is laminar and not turbulent flow. An algebraic equation relates the powers and products of variables, for instance: y^2 = 3x + 9x^2 A differential equation relates a function to its various derivatives, for example: x'' = -(k/m)x Is a differential equation with a general solution of x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t) In the case of freefall in an atmosphere, you normally will have some force of drag that is proportional to the velocity, and in the opposite direction, that means with air the differential equation is: F = ma = my'' = -mg -ky' or my'' + ky' = -mg y'' + (k/m)y' = -g Multiply this equation by e^(t*(k/m), and a little calculus dexterity shows d/dt(e^(t*(k/m))y') = -e^(t*(k/m)g By integrating both… [cont.]
Answered by supastremph - Wed Mar 11 03:58:59 2009
Q. Just wondering...Everytime our instructor has to stop and says "ignore air resistance that will require differential equations which you will not get into until your engineering and physics classes and besides its complicated stuff." What are differential equations really? I always see two variables piled up next to each other and one side equals to another side plus a constant. Differential measure the rate of change, I know that, but how precise is it if there's just a constant added to it each time
Asked by crazy girl - Wed Mar 11 03:45:27 2009 - - 2 Answers - 0 Comments
A. Only if there is laminar and not turbulent flow. An algebraic equation relates the powers and products of variables, for instance: y^2 = 3x + 9x^2 A differential equation relates a function to its various derivatives, for example: x'' = -(k/m)x Is a differential equation with a general solution of x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t) In the case of freefall in an atmosphere, you normally will have some force of drag that is proportional to the velocity, and in the opposite direction, that means with air the differential equation is: F = ma = my'' = -mg -ky' or my'' + ky' = -mg y'' + (k/m)y' = -g Multiply this equation by e^(t*(k/m), and a little calculus dexterity shows d/dt(e^(t*(k/m))y') = -e^(t*(k/m)g By integrating both… [cont.]
Answered by supastremph - Wed Mar 11 03:58:59 2009
How important is differential equations class?
Q. How important is differential equations class for chemical engineering major? Does the major rely a lot on this course and is it as important as multivariable calculus?? I wanted to take diff eq. during summer and was wondering if that was okay or should I take it during full semester...because over summer they don't go over details...are details important for future classes?
Asked by mka23 - Wed Jan 31 15:27:04 2007 - - 4 Answers - 0 Comments
A. The reason Newton invented calculus is to solve differential equations!! All your math classes up to this point are simply in preparation for solving differential equations. You might ask the prof who teaches it if there is any difference in the material covered in summer.
Answered by WildOtter - Wed Jan 31 17:48:56 2007
Q. How important is differential equations class for chemical engineering major? Does the major rely a lot on this course and is it as important as multivariable calculus?? I wanted to take diff eq. during summer and was wondering if that was okay or should I take it during full semester...because over summer they don't go over details...are details important for future classes?
Asked by mka23 - Wed Jan 31 15:27:04 2007 - - 4 Answers - 0 Comments
A. The reason Newton invented calculus is to solve differential equations!! All your math classes up to this point are simply in preparation for solving differential equations. You might ask the prof who teaches it if there is any difference in the material covered in summer.
Answered by WildOtter - Wed Jan 31 17:48:56 2007
Mathematica problem: How to solve coupled differential equations with no 'independent' variable?
Q. I am attempting to solve two coupled equations in two variables, say x and y. The problem is that one of the equations contains a term in dy/dx. There is no 'independent' variable per se. Both equations are quite nonlinear are require numerical solution. The NSolve and FindRoot functions will not work because of the differential while the Dsolve or NDsolve will not seem to work because of the lack of an 'independent variable to input. Any suggestions on how to approach this problem? I am attempting to solve two coupled equations in two variables, say x and y. The problem is that one of the equations contains a term in dy/dx. There is no 'independent' variable per se. Both equations are quite nonlinear are require numerical solution. The… [cont.]
Asked by locke9k - Tue Feb 6 11:09:50 2007 - - 1 Answers - 0 Comments
A. cud u post the equations
Answered by Maths Rocks - Tue Feb 6 11:15:15 2007
Q. I am attempting to solve two coupled equations in two variables, say x and y. The problem is that one of the equations contains a term in dy/dx. There is no 'independent' variable per se. Both equations are quite nonlinear are require numerical solution. The NSolve and FindRoot functions will not work because of the differential while the Dsolve or NDsolve will not seem to work because of the lack of an 'independent variable to input. Any suggestions on how to approach this problem? I am attempting to solve two coupled equations in two variables, say x and y. The problem is that one of the equations contains a term in dy/dx. There is no 'independent' variable per se. Both equations are quite nonlinear are require numerical solution. The… [cont.]
Asked by locke9k - Tue Feb 6 11:09:50 2007 - - 1 Answers - 0 Comments
A. cud u post the equations
Answered by Maths Rocks - Tue Feb 6 11:15:15 2007
How do you solve differential equations of the form dv/dt=f(t)+g(v)?
Q. The specific equation I'm trying to solve is: dv/dt = ae^(bt) - cv where a, b and c are positive constants. I know how to solve differential equations of the form dv/dt = f(t)g(v) but not dv/dt = f(t) + g(v).
Asked by Mark R - Fri Sep 7 13:42:35 2007 - - 2 Answers - 0 Comments
A. dv/dt + cv = ae^(bt) this is of the type dv/dt + P(t)v = Q(t) multiply through by e^( P(t)dt), in this case e^( cdt) = e^(ct) e^(ct).dv/dt + cv.dv/dt = ae^(bt).e^(ct) the LHS is then an exact derivative d/dt(ve^(ct)) = ae(bt + ct) integrating both sides ve^(ct) = ae(bt + ct)/(b + c) + K v = ae^(bt) / (b + c) + Ke(-ct)
Answered by fred - Sat Sep 8 09:33:53 2007
Q. The specific equation I'm trying to solve is: dv/dt = ae^(bt) - cv where a, b and c are positive constants. I know how to solve differential equations of the form dv/dt = f(t)g(v) but not dv/dt = f(t) + g(v).
Asked by Mark R - Fri Sep 7 13:42:35 2007 - - 2 Answers - 0 Comments
A. dv/dt + cv = ae^(bt) this is of the type dv/dt + P(t)v = Q(t) multiply through by e^( P(t)dt), in this case e^( cdt) = e^(ct) e^(ct).dv/dt + cv.dv/dt = ae^(bt).e^(ct) the LHS is then an exact derivative d/dt(ve^(ct)) = ae(bt + ct) integrating both sides ve^(ct) = ae(bt + ct)/(b + c) + K v = ae^(bt) / (b + c) + Ke(-ct)
Answered by fred - Sat Sep 8 09:33:53 2007
For differential equations, when the solution has repeated roots, why is a t added to the second solution?
Q. For the differential equation, second order, that ends up having repeated roots, why is a t added to the second assumed solution? For example, y=A*e^rt+Bt*e^rt. What is the reason behind choosing this?
Asked by unknown - Fri Sep 11 11:12:46 2009 - - 1 Answers - 0 Comments
A. Good question! 1) e^(rt) and e^(rt) most certainly are not linearly independent; so need to find a second root. 2) Why the extra t? Here's one way to get it. For a second order linear DE to have a double root, it may be written in the form y'' - 2ry' + r^2 y = 0 for some real r. (Note that the characteristic equation has double root r.) Since y = e^(rt) is one solution, let's use the change of variable y = e^(rt) * z, where z is also a function of t. [This is a standard change of variable trick for higher order linear DE's to reduce the order by 1.] So, y' = e^(rt) (z' + rz) and y'' = e^(rt)(z'' + 2rz' + r^2 z) by applying the product rule. Substituting this into the DE yields y'' - 2ry' + r^2 y = 0 ==> [e^(rt)(z'' + 2rz' + r^2 z) [cont.]
Answered by kb - Fri Sep 11 11:42:38 2009
Q. For the differential equation, second order, that ends up having repeated roots, why is a t added to the second assumed solution? For example, y=A*e^rt+Bt*e^rt. What is the reason behind choosing this?
Asked by unknown - Fri Sep 11 11:12:46 2009 - - 1 Answers - 0 Comments
A. Good question! 1) e^(rt) and e^(rt) most certainly are not linearly independent; so need to find a second root. 2) Why the extra t? Here's one way to get it. For a second order linear DE to have a double root, it may be written in the form y'' - 2ry' + r^2 y = 0 for some real r. (Note that the characteristic equation has double root r.) Since y = e^(rt) is one solution, let's use the change of variable y = e^(rt) * z, where z is also a function of t. [This is a standard change of variable trick for higher order linear DE's to reduce the order by 1.] So, y' = e^(rt) (z' + rz) and y'' = e^(rt)(z'' + 2rz' + r^2 z) by applying the product rule. Substituting this into the DE yields y'' - 2ry' + r^2 y = 0 ==> [e^(rt)(z'' + 2rz' + r^2 z) [cont.]
Answered by kb - Fri Sep 11 11:42:38 2009
Where can I find the book "Differential equations with boundary value problems?
Q. I am an engineering undergrad; I want the following book for free download: Differential equations with boundary value problems - Dennis G. Zill, Michael Cullen Any help will be appreciated, Thanks I said "free download". I am in a location where buying from amazon will be to expensive sense it would involve currency conversion. So please if you have any information about a site where I can download this book for free, let me know.
Asked by Taimy_22 - Wed Feb 11 09:50:30 2009 - - 2 Answers - 0 Comments
A. hey mate, whilst amazon is good online bookstore - thankfully there is a far better (and by better I mean cheaper) online store www.abebooks.com I would probably purchase 80% of my texts from here, and even with the transportation cost from the US to AUS (cheap as chips if your located in the US) it is remarkably cheap. Generally I pay roughly $5 - $20 AUD per book (oh I should have mentioned - its a second hand online bookstore) So unless you want a fresh new copy, I would thoroughly recommend it. Hope this helps, David
Answered by unknown - Wed Feb 11 10:02:58 2009
Q. I am an engineering undergrad; I want the following book for free download: Differential equations with boundary value problems - Dennis G. Zill, Michael Cullen Any help will be appreciated, Thanks I said "free download". I am in a location where buying from amazon will be to expensive sense it would involve currency conversion. So please if you have any information about a site where I can download this book for free, let me know.
Asked by Taimy_22 - Wed Feb 11 09:50:30 2009 - - 2 Answers - 0 Comments
A. hey mate, whilst amazon is good online bookstore - thankfully there is a far better (and by better I mean cheaper) online store www.abebooks.com I would probably purchase 80% of my texts from here, and even with the transportation cost from the US to AUS (cheap as chips if your located in the US) it is remarkably cheap. Generally I pay roughly $5 - $20 AUD per book (oh I should have mentioned - its a second hand online bookstore) So unless you want a fresh new copy, I would thoroughly recommend it. Hope this helps, David
Answered by unknown - Wed Feb 11 10:02:58 2009
Would you help with a question about differential equations?
Q. Salut I was wondering if you could tell me what the difference between a "stiff" and a "nonstiff" differential equation is? I am not too lazy to research the answer myself (this question is in fact part of said research), I am just new to the discipline and, never having had a formal course, seek information where I can find it. Thanks for reading my senseless blithering, Max
Asked by Max - Sun Aug 24 12:18:26 2008 - - 1 Answers - 0 Comments
A. Like many things in life, the best source is Wikipedia [1]. To make a long story short: a stiff equation is one that is unstable with respect to numerical schemes, where a nonstiff equation is the opposite. To see the picture better, we first consider what it means to solve a differential equation numerically. Basically, since the computer is a digital device, it is not advisable to try to model the entire continuum; rather, a numerical scheme to solve a differential equation starts by chosing a grid on which to solve the equation. Once we have a grid, we can reduce the differential equation to relations between values on a finite number of points. Now, depending on the spacing between grid elements that one choses, one may sometimes end… [cont.]
Answered by jaz_will - Mon Aug 25 20:33:11 2008
Q. Salut I was wondering if you could tell me what the difference between a "stiff" and a "nonstiff" differential equation is? I am not too lazy to research the answer myself (this question is in fact part of said research), I am just new to the discipline and, never having had a formal course, seek information where I can find it. Thanks for reading my senseless blithering, Max
Asked by Max - Sun Aug 24 12:18:26 2008 - - 1 Answers - 0 Comments
A. Like many things in life, the best source is Wikipedia [1]. To make a long story short: a stiff equation is one that is unstable with respect to numerical schemes, where a nonstiff equation is the opposite. To see the picture better, we first consider what it means to solve a differential equation numerically. Basically, since the computer is a digital device, it is not advisable to try to model the entire continuum; rather, a numerical scheme to solve a differential equation starts by chosing a grid on which to solve the equation. Once we have a grid, we can reduce the differential equation to relations between values on a finite number of points. Now, depending on the spacing between grid elements that one choses, one may sometimes end… [cont.]
Answered by jaz_will - Mon Aug 25 20:33:11 2008
how do i explain partial differential equations to a fifth grader?
Q. i have to write a children's book on sonya kovalevsky who discover some theorem on partial differential equations any help is very much appreciated
Asked by phatty34 - Fri Mar 20 21:17:25 2009 - - 3 Answers - 0 Comments
A. Partial differential equations is a subject which is taught at the university level, and only to students in 2nd or 3rd year mathematics. It is completely beyond any hope of being understood by someone ( * ) who is still working on the intricacies of long division, or maybe multiplication of fractions ! The best you can hope to do is explain some of the concepts in the most general terms and hope they get the idea that this work was somehow important. It is also a good idea to watch out for making it seem like some impossibly difficult thing, lest you poison their young minds against mathematics without their ever having had a chance to get started. It is a fine line to tread. ( * - geniuses like a young Gauss, excepted )
Answered by MathMan TG - Sat Mar 21 04:22:56 2009
Q. i have to write a children's book on sonya kovalevsky who discover some theorem on partial differential equations any help is very much appreciated
Asked by phatty34 - Fri Mar 20 21:17:25 2009 - - 3 Answers - 0 Comments
A. Partial differential equations is a subject which is taught at the university level, and only to students in 2nd or 3rd year mathematics. It is completely beyond any hope of being understood by someone ( * ) who is still working on the intricacies of long division, or maybe multiplication of fractions ! The best you can hope to do is explain some of the concepts in the most general terms and hope they get the idea that this work was somehow important. It is also a good idea to watch out for making it seem like some impossibly difficult thing, lest you poison their young minds against mathematics without their ever having had a chance to get started. It is a fine line to tread. ( * - geniuses like a young Gauss, excepted )
Answered by MathMan TG - Sat Mar 21 04:22:56 2009
What are differential equations and how t do you solve it?
Q. I just don't get what differntial equations are and I don't know how to solve it. Please don't use complex terms in explaining.
Asked by Mario - Wed Oct 7 17:00:38 2009 - - 2 Answers - 0 Comments
A. differential equations are equations that deals with dy and dx or even in higher forms like d^ny and dx^n. it's sort of like calculus but an extention of it. remember unibomber the professor from Berkley. he specialize in solving it. there is a whole book written about it. it mainly deal with e^x because when u differentiate e^x, it's equal to e^x itself. so solution is easily find in some form of exponential. to go further, u start to solve using Laplace transform. Laplace transform turns the differential equations into algebraic accounting. and in partial differnentiation equations, solutions are found in Fourier series using sin and cosine. there is a characteristic solution plus a particular solution. u can't learn to solve… [cont.]
Answered by Fred Osim - Wed Oct 7 17:37:21 2009
Q. I just don't get what differntial equations are and I don't know how to solve it. Please don't use complex terms in explaining.
Asked by Mario - Wed Oct 7 17:00:38 2009 - - 2 Answers - 0 Comments
A. differential equations are equations that deals with dy and dx or even in higher forms like d^ny and dx^n. it's sort of like calculus but an extention of it. remember unibomber the professor from Berkley. he specialize in solving it. there is a whole book written about it. it mainly deal with e^x because when u differentiate e^x, it's equal to e^x itself. so solution is easily find in some form of exponential. to go further, u start to solve using Laplace transform. Laplace transform turns the differential equations into algebraic accounting. and in partial differnentiation equations, solutions are found in Fourier series using sin and cosine. there is a characteristic solution plus a particular solution. u can't learn to solve… [cont.]
Answered by Fred Osim - Wed Oct 7 17:37:21 2009
Calculus question based on Seperable Differential Equations?
Q. I know that this problem involves seperable differential equations, but I have no idea how to set it up! Let C(t) be the number of cougars on an island at time t years (where t is greater than or equal to zero). The number of cougars is increasing at a rate directly proportional to 3500-C(t). Also, C(0)=1000, and C(5)=2000. Find C(t) as a function of t only.
Asked by unknown - Sat May 23 14:47:40 2009 - - 2 Answers - 0 Comments
A. n=3500-at,because you say proportional 1000=3500-0 x t Something is wrong with the problem.
Answered by nozar nazari - Sat May 23 14:56:36 2009
Q. I know that this problem involves seperable differential equations, but I have no idea how to set it up! Let C(t) be the number of cougars on an island at time t years (where t is greater than or equal to zero). The number of cougars is increasing at a rate directly proportional to 3500-C(t). Also, C(0)=1000, and C(5)=2000. Find C(t) as a function of t only.
Asked by unknown - Sat May 23 14:47:40 2009 - - 2 Answers - 0 Comments
A. n=3500-at,because you say proportional 1000=3500-0 x t Something is wrong with the problem.
Answered by nozar nazari - Sat May 23 14:56:36 2009
Can i teach myself Differential Equations?
Q. How difficult are Differential Equations? Using a textbook and the internet, is it reasonable to be able to become proficcient in 3 months? I have a year of calculus and physics under my belt . Thanks everybody!! I learn as well from textbooks as from lectures. My calculus takes me into advanced integration and starting multivariate calculus (calc III).
Asked by bodicus - Thu Mar 29 23:22:33 2007 - - 7 Answers - 0 Comments
A. Depends. What's your learning style, and what's your work ethic like? For me, I learn math best by simply studying the textbook on my own. Math classes help me impose discipline on myself (in terms of getting assignments done by some deadline), but I usually learn very little from the lectures themselves, and I hardly ever have questions to ask the professor. If you have a similar style, then sure, you should be able to learn differential equations on your own. On the other hand, if you need regular feedback from a teacher or other students, then it could be a very different story. In any case, you'll need to be sure you have the prerequisite knowledge. It's been a looong time since I had differential equations, but it came somewhere… [cont.]
Answered by Bramblyspam - Thu Mar 29 23:31:56 2007
Q. How difficult are Differential Equations? Using a textbook and the internet, is it reasonable to be able to become proficcient in 3 months? I have a year of calculus and physics under my belt . Thanks everybody!! I learn as well from textbooks as from lectures. My calculus takes me into advanced integration and starting multivariate calculus (calc III).
Asked by bodicus - Thu Mar 29 23:22:33 2007 - - 7 Answers - 0 Comments
A. Depends. What's your learning style, and what's your work ethic like? For me, I learn math best by simply studying the textbook on my own. Math classes help me impose discipline on myself (in terms of getting assignments done by some deadline), but I usually learn very little from the lectures themselves, and I hardly ever have questions to ask the professor. If you have a similar style, then sure, you should be able to learn differential equations on your own. On the other hand, if you need regular feedback from a teacher or other students, then it could be a very different story. In any case, you'll need to be sure you have the prerequisite knowledge. It's been a looong time since I had differential equations, but it came somewhere… [cont.]
Answered by Bramblyspam - Thu Mar 29 23:31:56 2007
Eigenvalues and eigenvectors for linear differential equations?
Q. Find all possible Eigenvalues and eigenvectors for linear differential equations for the following differential equations: x' = 5x + (8/3)y - (2/3)z y' = 2x + (2/3)y + (4/3)z z' = -4x - (4/3)y - (8/3)z
Asked by Harsh Saini - Sun Apr 5 20:13:17 2009 - - 1 Answers - 0 Comments
A. Hey, I don't really feel like doing a bunch of menial calculations so I just popped the matrix into Matlab. I found the following eigenvalue/vector pairs: 6 - [4,1,-2]/sqrt(21) -3 - [1,-2,4]/sqrt(21) 0 - [-2,4,1]/sqrt(21) You can get rid of the sqrt(21) if you don't need your eigenvectors to be orthonormal. You could find this by hand by solving the characteristic polynomail det(A-kI)=0, but that's a silly amount of work for this problem. At least in my opinion. Good luck.
Answered by Michael M - Mon Apr 6 13:23:04 2009
Q. Find all possible Eigenvalues and eigenvectors for linear differential equations for the following differential equations: x' = 5x + (8/3)y - (2/3)z y' = 2x + (2/3)y + (4/3)z z' = -4x - (4/3)y - (8/3)z
Asked by Harsh Saini - Sun Apr 5 20:13:17 2009 - - 1 Answers - 0 Comments
A. Hey, I don't really feel like doing a bunch of menial calculations so I just popped the matrix into Matlab. I found the following eigenvalue/vector pairs: 6 - [4,1,-2]/sqrt(21) -3 - [1,-2,4]/sqrt(21) 0 - [-2,4,1]/sqrt(21) You can get rid of the sqrt(21) if you don't need your eigenvectors to be orthonormal. You could find this by hand by solving the characteristic polynomail det(A-kI)=0, but that's a silly amount of work for this problem. At least in my opinion. Good luck.
Answered by Michael M - Mon Apr 6 13:23:04 2009
Differential equations question concerning compound interest?
Q. A deposit is made to a bank account paying 8% interest compounded continuously. Payments are made from this account at a rate of $5000 per year. (a) Write a differential equation for the balance, B, in the account after t years (b) Write the solution to the differential equation. Use A as your unknown constant, where A is the coefficient of your exponential term.
Asked by wildcat11 - Wed May 7 03:15:25 2008 - - 1 Answers - 0 Comments
A. B' = B*0.08 -5000 B= A exp(0.08t) - 5000/0.08
Answered by hustolemyname - Wed May 7 03:23:23 2008
Q. A deposit is made to a bank account paying 8% interest compounded continuously. Payments are made from this account at a rate of $5000 per year. (a) Write a differential equation for the balance, B, in the account after t years (b) Write the solution to the differential equation. Use A as your unknown constant, where A is the coefficient of your exponential term.
Asked by wildcat11 - Wed May 7 03:15:25 2008 - - 1 Answers - 0 Comments
A. B' = B*0.08 -5000 B= A exp(0.08t) - 5000/0.08
Answered by hustolemyname - Wed May 7 03:23:23 2008
Is textbook Differential Equations with Boundary-Value Problems by Dennis Zill a good one?
Q. I intend to buy this for math self-learning. Here is the one : Anybody use it? Thanks!
Asked by gogo - Mon Jan 26 09:08:32 2009 - - 2 Answers - 0 Comments
A. I haven't used that text, but the one I used for my DE class was the following one: It actually happens to be cheaper than the one your thinking of buying. It was a great text, not only does it cover techniques for soling DE's, but also goes into the theory. Hope this helps! p.s. I guess its something you'd probably want to know, I taught myself the entire course using that textbook actually since my professor really sucked a teaching. So its definitely a great text for self-learning. If you don't mind downloading texts "illegally" then you could probably find any text you're looking for here: You just have to make sure you have the proper viewer to be able to see the text you download.
Answered by Buri - Mon Jan 26 09:18:42 2009
Q. I intend to buy this for math self-learning. Here is the one : Anybody use it? Thanks!
Asked by gogo - Mon Jan 26 09:08:32 2009 - - 2 Answers - 0 Comments
A. I haven't used that text, but the one I used for my DE class was the following one: It actually happens to be cheaper than the one your thinking of buying. It was a great text, not only does it cover techniques for soling DE's, but also goes into the theory. Hope this helps! p.s. I guess its something you'd probably want to know, I taught myself the entire course using that textbook actually since my professor really sucked a teaching. So its definitely a great text for self-learning. If you don't mind downloading texts "illegally" then you could probably find any text you're looking for here: You just have to make sure you have the proper viewer to be able to see the text you download.
Answered by Buri - Mon Jan 26 09:18:42 2009
How do you solve problems involving applied differential equations?
Q. I'm completely confused? What exactly are you suppose to do. How exactly are you suppose to make an equation out of them??? An object is dropped from a great height. Its acceleration due to gravity is 9.8 meters per second per second and its deceleration due to air resistance is proportional to its velocity. After a long period of time, its terminal velocity of 30 meters per second is achieved. What was its velocity after 10 seconds? Thanks.
Asked by KB - Fri Jul 31 19:11:53 2009 - - 1 Answers - 0 Comments
A. Acceleration is the time derivative of velocity. Take downward as positive. Then the acceleration a(t) is given by a(t) = dv(t)/dt = 9.8 - kv(t) where k is the proportionality constant for air resistance. Terminal velocity is reached when there is no net force on the falling object. This never really happens, but for long times it is arbitrarily closely achieved. Since force = mass*acceleration, this means that the acceleration is effectively zero. Thus, for terminal velocity of 30m/s, we have 9.8 - k(30) = 0 Solving for k, k = 9.8/30 so solve dv/dt = 9.8 - (9.8/30)v subject to the initial condition v(0) = 0 (it was dropped, not thrown) Note that the equation may be written dv/dt + (9.8/30)v = 9.8 This is first order linear… [cont.]
Answered by Ron W - Fri Jul 31 20:55:16 2009
Q. I'm completely confused? What exactly are you suppose to do. How exactly are you suppose to make an equation out of them??? An object is dropped from a great height. Its acceleration due to gravity is 9.8 meters per second per second and its deceleration due to air resistance is proportional to its velocity. After a long period of time, its terminal velocity of 30 meters per second is achieved. What was its velocity after 10 seconds? Thanks.
Asked by KB - Fri Jul 31 19:11:53 2009 - - 1 Answers - 0 Comments
A. Acceleration is the time derivative of velocity. Take downward as positive. Then the acceleration a(t) is given by a(t) = dv(t)/dt = 9.8 - kv(t) where k is the proportionality constant for air resistance. Terminal velocity is reached when there is no net force on the falling object. This never really happens, but for long times it is arbitrarily closely achieved. Since force = mass*acceleration, this means that the acceleration is effectively zero. Thus, for terminal velocity of 30m/s, we have 9.8 - k(30) = 0 Solving for k, k = 9.8/30 so solve dv/dt = 9.8 - (9.8/30)v subject to the initial condition v(0) = 0 (it was dropped, not thrown) Note that the equation may be written dv/dt + (9.8/30)v = 9.8 This is first order linear… [cont.]
Answered by Ron W - Fri Jul 31 20:55:16 2009
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Aryth
Fri, 06 Nov 2009 01:57:15 GM
(a) Using the separation of variables technique, and letting the separation constant be denoted E (which turns out to be the total energy of the.
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