Homogeneous Linear Differential Equations
Date: 05/06/2001 at 11:29:06 From: Tim Subject: Solving linear differential equations Hi, I have a couple of questions on solving differential equations of higher order. (note: the " ' " in y' stands for the first derivative of y(x). I do know calculus. How would one solve y''-2y'+5y = 0 ? y''-2y'+5y = (e^x)+1 I have looked in a book that solved y''+ y = 0 by changing it into r^2 + 1 = 0 (I know the square of a number and the second derivative are quite different) by solving for r, which is +- i (imaginary number) then because r = +- i, y = (c1)cos(x) +(c2)sin(x), and c1 and c2 are probably constants. Also, I saw that if the equation could be factored: ex. y''+5y'+6y = 0 "factors" into (r+3)(r+2) = 0 where roots r = -3, -2 thuse y = (c1)e^(-3x)+(c2)e^(-2x) because the r is integral. Are there any general rules for solving such differential equations involving y'''', y''', y'', y', and y? Also, how would you solve (y''')^2+(y'')^3-y = 0? (the third derivative of y squared + the second derivative of y cubed - y = 0) ? Can you help me? Thanks!
Date: 05/09/2001 at 16:08:41 From: Doctor Jordi Subject: Re: Solving linear differential equations Hi, Tim - thank you for writing. You pose very interesting questions! There is no general method for solving differential equations (at least, I don't think anyone has found one, and it doesn't seem very likely anyone will find one any time within the next couple of centuries). A few very select equations can be solved explicitly (explicitly meaning that one can give an answer in the form of a formula that describes a function), like the ones you have seen in your books, others have answers in the form of a series expansion, (e.g. Taylor polynomials), and a bit broader range of differential equations can be approximated numerically (for examples of such approximations look up Euler's method or Runge-Kutta methods, both of which should appear in any decent differential equation textbook). However, this means that we do not know yet how to solve the majority of differential equations that we can imagine. Let us have a look at the equations you proposed. y" - 2y' + 5y = 0 (1) y" - 2y' + 5y = e^x + 1 (2) y" + 5y' + 6y = 0 (3) (y"')^2 + (y")^3 - y = 0 (4) The first step in solving a differential equation is to classify it. Then, after the type of equation is recognized, one may attempt to solve it by very specific methods. In the equations above, I will assume that y is the dependent variable and that the independent variable is x. Equations (1) and (3) are called constant-coefficient linear second- order homogeneous ordinary differential equations. What a mouthful! Let's see what each of these words means individually: equation - An equation is a statement which claims that two quantities are equal. differential - In particular, a differential equation involves a function (e.g. y(x)) and its derivatives.. ordinary - An ordinary differential equation (ODE) only involves ordinary derivatives, as opposed to a partial differential equations which involve partial derivatives. linear - In a linear ODE, the functions and its derivatives are only added together and perhaps multiplied by functions of the independent variable, and nothing else. That said, an ODE such as (y")^2 + y = x is not a linear equation, since the second derivative of y is being raised to a power. Equations (1), (2), and (3), however, are all linear. second-order - In general, the order of an equation is the same as the order of the highest derivative in it. In particular, equation (1) has order two, since its highest derivative is y". homogeneous - A linear ODE is said to be homogeneous if the function of the independent variable that is not multiplied by any of the derivatives of the dependent variable is 0. For example, the equation (1) and (3) are homogeneous, while equation (2) is not. constant-coefficient - A linear ODE may have derivatives of the dependent variable being multiplied by functions of the independent variable. When these functions are constants (e.g. 2, 0, pi, -i), then the equation is called a constant-coefficient equation. Equations (1), (2), and (3) are constant-coefficient linear ODEs. We must be careful about how we solve equations such as (1) and (3). We are not "changing" the equation into a quadratic. We are solving an auxiliary equation that tells us something about the corresponding differential equation. Do you know where the auxiliary equation comes from? What it means? It's motivated by a clever trick (basically all solutions to differential equations are motivated by tricks). We first notice that in equations such as (1) and (3) we can express the second derivative as linear combinations of the first and second derivatives: y" = 2y' - 5y This suggests that a function twice differentiated is the same as the function itself times a constant plus its derivative times a constant. By the "method" of judicious guessing, we choose an exponential function of the form e^(rx) where r is a constant, because we know that these functions have properties which seem to fit equation (1). Let y(x) = e^(rx) so that y'(x) = re^(rx) and y"(x) = r^2*e^(rx) Substituting these values of y' and y" back into equation (1) we observe that r^2*e^(rx) - 2re^(rx) + 5e^(rx) = 0 Or, after factoring e^(rx) e^(rx)[r^2 - 2r + 5] = 0 Since the exponential function is always positive, it follows that the above equation is satisfied if and only if r^2 - 2r + 5 = 0 Which is the auxiliary equation you are familiar with. From that equation we can find that r = 1 +- 2i. This gives two linearly independent [SEE NOTE 1 AT THE END OF THIS ANSWER] solutions to (1) of the form y1(x) = e^[(1 + 2i)x] and y2(x) = e^[(1 - 2i)x] Now, although this next step requires a bit of justification (I recommend you read your textbook for this part), you and I can be a little loose and apply Euler's formula, which works for exponentials with complex powers. Euler's formula says that e^(a + bi) = (e^a)*[cos(b) + i*sin(b)] From which we can express y1 and y2 as y1 = (e^x)*[cos(2x) + i*sin(2x)] and y2 = e*^(x)[cos(2x) - i*sin(2x)] Now, since the general solution of equation (1) is expressible as Ys = c1*y1 + c2*y2 Ys = (c1)(e^x)[cos(2x) + i*sin(2x)] + (c2)(e^x)[cos(2x) + i*sin(2x)] where c1 and c2 are any complex scalars, we can express it as Ys = (k1)(e^x)cos(2x) + (k2)*(e^x)sin(2x) [SEE NOTE 2 AT THE END OF THIS ANSWER.] Where k1 and k2 are any other scalars, not necessarily complex. We can choose k1 and k2 in such a way as to form any of all the possible real solutions to (1). This gives the complete solution to equation (1). Equation (2) is the same as (1), except that (2) is nonhomogeneous. Thankfully, we have already done the work for solving (2) when we solved (1). Say there was another solution for (2), Yp, which we call the _particular solution_ of the equation. We have found Yg above. Now, let Yg = Ys + Yp. Look at what happens when we substitute this into equation (2): (Yg)" - 2(Yg)' + 5(Yg) = e^x + 1 (Ys + Yp)" - 2(Ys + Yp)' + 5(Ys + Yp) = e^x + 1 (Ys" - 2Ys' + 5Ys) + (Yp" - 2Yp' + 5Yp) = e^x +1 ^ ^ | | This quantity = 0 This quantity = e^x + 1 since we have proved it since we postulated that Yp solved in equation (1) equation (2) So that if we could find Yp, we could find the set of all possible solutions to equation (2). [SEE NOTE 3 AT THE BOTTOM OF THIS ANSWER.] Thus we seek Yp, which solves equation (2). There are several methods we could try to employ. I will use one I personally like due to its simplicity; what it does is "guess" a solution of the form Yp = Ae^x + B and then find the values of A and B that satisfy the equation. This is called the method of undetermined coefficients. (Ae^x + B)" - 2(Ae^x + B)' + 5(Ae^x + B) = e^x + 1 e^x(A - 2A + 5A) + 5B = e^x + 1 So on equating coefficients we see that 4A = 1 5B = 1 gives the particular solution Yp = (1/4)e^x + 1/5 Finally, by the previous arguments, all possible solutions of equation (2) are of the form Yg = (k1)(e^x)cos(2x) + (k2)(e^x)sin(2x) + (1/4)e^x + 1/5 Where k1 and k2 can again be any constants (not necessarily complex). Equation (3) can be solved by methods just like the ones I described above. Try to do it on your own. I tried solving equation (4) on a computer and got a very ugly output. If it has a solution, it's a solution far too complex to be interesting. Again, I must emphasize that only very few and very specific differential equations have solutions. The theory of differential equations is very rich and very varied; complex problems abound, many unsolved as of yet. Finally, about your question for solving differential equations of higher order, I'll let you answer that one yourself. Look at the arguments I employed in this answer for solving the second-order equations you proposed. Is there anything special about my arguments that would make them apply only to constant-coefficient linear differential equations of order 2? Would they apply to constant- coefficient linear differential equations of order 3? How about order 42? Do all equations of the form of (1), (2) and (3) of any order always have solutions? How many linearly independent solutions will they have? How about the fundamental theorem of algebra, does that have an application in this case? I hope all this helped. Please write back if you would like to talk more about this, or if you have more questions. A couple of remarks concerning the theory of differential equations... NOTE 1: When we are solving a differential equation such as (1), we are looking for two linearly independent solutions from which we can form all possible solutions of the system. In the case of just two solutions, "linearly independent" means that they are not constant multiples of each other. That is, if y1 and y2 are linearly independent, then there is no constant c such that c*y1 = y2. NOTE 2: Sometimes, such as in this case, the auxiliary equation has complex solutions, but we are usually interested only in the real solutions of the equation, because our differential equation usually is a model describing a physical phenomenom (equation (1) could be modelling a mass attached to a spring that is oscillating and being driven by a force). That is why, once we have formed the general solution, y, which may contain complex coefficients, we can still write the same solution using two different linearly independent solutions without using complex numbers. For more details on how this is done and why it works, I suggest that you consult your textbook (or write back, if you need more help). NOTE 3: When we are solving a non-homogeneous equation, we only need, in fact, ONE particular solution, Yp, and the general solution Ys to the corresponding homogeneous system. It can be shown (and perhaps you may read about it in your textbook as well) that all possible solutions of the non-homogeneous equation will be of the form Ys + Yp, where Ys and Yp are the general solution of the homogeneous equation and a particular solution of the nonhomogeneous equation respectively. - Doctor Jordi, The Math Forum http://mathforum.org/dr.math/
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