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PERTEMUAN VI. Penyelesaian Khusus (Particular Solution ) Nonhomogeneous Linear Equations Consider the nonhomogeneous linear equation We have seen that the general solution is given by , where is a particular solution and is the general solution of the associated homogeneous equation. We will not discuss the case of non-constant coefficients. Therefore, we will restrict ourself only to the following type of equation: Using the previous section, we will discuss how to find the general solution of the associated homogeneous equation Therefore, the only remaining obstacle is to find a particular solution to (NH). In the second order differential equations case, we learned the two methods: Undetermined Coefficients Method and the Variation of Parameters . These VI-1 two methods are still valid in the general case, but the second one is very hard to carry. Undetermined Coefficients Method Example: Find a particular solution of Solution: Let us follow these steps: (1) Characteristic equation We have the factorization . Therefore, the roots are 0, 2, -2 and they are all simple. (2) We have to split the equation into the following two equations: ; (3) The particular solution to the equation (1): (3.1) We have which is a simple root. Then s = 1; (3.2) The guessed form for the particular solution is , where A and B are to be determined. We will omit the detail of the calculations. We get A = -1/8 and B = 0. Therefore, we have VI-2 ; (4) The particular solution to the equation (2): (4.1) We have which is not a root. Then s = 0; (4.2) The guessed form for the particular solution is where A and B are to be determined. We will omit the detail of the calculations. We get A = 0 and B=-3/5. Therefore, we have ; (5) The particular solution to the original equation is given by (4) The particular solution to the equation (2): (4.1) We have which is not a root. Then s = 0; (4.2) The guessed form for the particular solution is where A and B are to be determined. We will omit the detail of the calculations. We get A = 0 and B=-3/5. Therefore, we have ; (5) The particular solution to the original equation is given by VI-3 Method of Variation of Parameters This method has no prior conditions to be satisfied. Therefore, it may sound more general than the previous method. We will see that this method depends on integration while the previous one is purely algebraic which, for some at least, is an advantage. Consider the equation In order to use the method of variation of parameters we need to know that is a set of fundamental solutions of the associated homogeneous equation y'' + p(x)y' + q(x)y = 0. We know that, in this case, the general solution of the associated homogeneous equation is . The idea behind the method of variation of parameters is to look for a particular solution such as where and are functions. From this, the method got its name. The functions and are solutions to the system , which implies , VI-4 where is the wronskian of and . Therefore, we have Summary: Let us summarize the steps to follow in applying this method: (1) Find a set of fundamental solutions of the associated homogeneous equation y'' + p(x)y' + q(x)y = 0. (2) Write down the form of the particular solution ; (3) Write down the system ; (4) Solve it. That is, find (5) Plug and and ; into the equation giving the particular solution. Example: Find the particular solution to Solution: Let us follow the steps: (1) A set of fundamental solutions of the equation y'' + y = 0 is ; (2) The particular solution is given as VI-5 (3) We have the system ; (4) We solve for and , and get Using techniques of integration, we get (5) The particular solution is: or Remark: Note that since the equation is linear, we may still split if necessary. For example, we may split the equation , into the two equations then, find the particular solutions for (1) and particular solution for the original equation by for (2), to generate a VI-6 Metode Singkat ( Dn + bn-1 Dn-1 + b0D ) y = g(x) F(D) = g(x) 1. Bila g(x) = e Maka yp = ax 1 g ( x) D bn 1 D n 1 , , , b0 D n 1 1 ax e ax e , F ( D) F (a) = 1 e ax , F (a) Bila F(a) = 0 maka yp = x Bila F’(a) = 0 maka yp = x 2 2. Bila 1 e ax , F " (a) g(x) = xn Maka = ' yp = 1 g ( x) D bn 1 D n 1 , , , b0 D n 1 diuraikan F ( D) 1 xn F ( D) 3. Bila g(x) = sin ax Maka = yp = atau cos ax 1 g ( x) D bn 1 D n 1 , , , b0 D n 1 sin ax F (D 2 ) VI-7 = 1 sin ax F (a 2 ) 4. Bila g(x) = eax φ(x) Maka yp = eax 1 ( x) = F ( D a) 1 ( x) F ( D a) e-a x ∫e-ax φ (x) dx Example: Find a particular solution to the equation Solution: Let us follow these steps: (1) First, we notice that the conditions are satisfied to invoke the method of undetermined coefficients. (2) We split the equation into the following three equations : (2) The root of the characteristic equation and r = 4. are r = -1 Particular solution to Equation (1): Since , and , then , which is not one of the roots. Then s = 0. The particular solution is given as VI-8 If we plug it into the equation (1), we get , which implies A = -1/2, that is, Particular solution to Equation (2): Since , and β = 1, then α + βi = i , which is not one of the roots. Then s = 0. The particular solution is given as If we plug it into the equation (2), we get , which implies Particular solution to Equation (3): Since α = -1, and β = 0 , then α + βi = -1 which is one of the roots. Then s = 1. VI-9 The particular solution is given as If we plug it into the equation (3), we get , which implies , that is (5) A particular solution to the original equation is Example 3. Apply the First Shift Theorem to the following differential equations to determine the particular integral for each: a. b. c. Example 4. Give the general solutions to the following differential equations: a. y + 4y + 3y = e-2x; b. y + 4y + 3y = 5 cos(2x); VI-10 Example 5. Solve the following differential equation: y + 4y + 4y = 24 sin(2x), given that y(0) = 1 and y'(0) = 2. Solucia The menu of this page is in simple mode. There is a menu in expert mode for more options. The expert menu enable to get some animated curves of solutions. y' + 2*y = x+4 Show Resolution of the differential equation : General solution. Cauchy's solution. In this case , give the initial value(s) below , separate with comma : I.C. : . Curves representations: initial values of "I.C.") , in the interval from general solutions to In option , amplitude : from one solution (with . to . To know more , click here . Solucia Attention,a differential equation of order 1 means 1 initial condition. But you have gave 0 of those. VI-11 The general solution is : To obtain a particular solution of the equation, It's missing the initial value. Here is the evolution of the curve according to the missing value : TERIMA KASIH VI-12