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Pertemuan V Nonlinear Second Order Differential Equations In general, little is known about nonlinear second order differential equations , but two cases are worthy of discussion: (1) Equations with the y missing Let v = y'. Then the new equation satisfied by v is This is a first order differential equation. Once v is found its integration gives the function y. Example 1: Find the solution of Solution: Since y is missing, set v=y'. Then, we have This is a first order linear differential equation. Its resolution gives Since v(1) = 1, we get V-1 Consequently, we have Since y' = v, we obtain the following equation after integration The condition y(1) = 2 gives Therefore, we have Note that this solution is defined for x > 0. (2) Equations with the x missing Let v = y'. Since we get V-2 This is again a first order differential equation. Once v is found then we can get y through which is a separable equation. Beware of the constants solutions. Example 2: Find the general solution of the equation Solution: Since the variable x is missing, set v=y'. The formulas above lead to This a first order separable differential equation. Its resolution gives Since we get y' = 0 or Since this is a separable first order differential equation, we get, after resolution, , V-3 where C and are two constants. All the solutions of our initial equation are Note that we should pay special attention to the constant solutions when solving any separable equation. This may be source of mistakes... Homogeneous Linear Equations with Constant Coefficients A second order homogeneous equation with constant coefficients is written as where a, b and c are constant. This type of equation is very useful in many applied problems (physics, electrical engineering, etc..). Let us summarize the steps to follow in order to find the general solution: (1) Write down the characteristic equation This is a quadratic equation. Let r1 and r2 be its roots ( we have ; (2) If r1 and r2 are distinct real numbers (this happens if b2 – 4 ac > 0 ), then the general solution is V-4 (3) If r1 and r2 (which happens if b2 – 4ac = 0 ), then the general solution is (4) If r1 and r2 are complex numbers (which happens if b2 – 4 ac < 0 ), then the general solution is where , that is, Example: Find the solution to the IVP Solution: Let us follow the steps: 1 Characteristic equation and its roots V-5 Since 4-8 = -4<0, we have complex roots . Therefore, and ; 2 General solution ; 3 In order to find the particular solution we use the initial conditions to determine and . First, we have . Since , we get From these two equations we get , which implies V-6 Summary Equation: ay Auxiliary equation: ak2 + bk + c = 0 Solutions to the Auxiliary Equation General Solution Distinct real roots k1, k2 Equal real roots k1 = k2 (= k) Complex roots k1 = A + iB and k2 = A - iB y = eA[Pcos(Bx) + Qsin(Bx)] Example 1. Find the general solution of the following D.E.'s: a. y” + 4y’ + 3y = 0 b. y” + 4y’ + 4y = 0 c. y” + 4y’ + 5y = 0 The two constants appearing in the complementary function are usually determined by knowing two conditions associated with the problem. If the conditions both apply at the same value of x, (usually x = 0), the problem is called an initial value problem. Example 2. Solve the initial value problem y + 4y + 3y = 0 given y = 1 and y = 3 when x = 0. Example: Find the general solution of Solution: Let us follow these steps: (1) Characteristic equation V-7 Its roots are the complex numbers In the analytical form, these roots are ; (2) Independent set of solutions (2.1) The complex roots and generate the two solutions ; (2.2) The complex roots generate the two solutions ; (3) The general solution is V-8 Soal – soal 1. y″ - 3y′ + 2 y = 0 2. y” + y’ – 30 y = 0 ans . y = C1 ex + C2 e2x 3. y”” – 4 y’” + 8 y” – 8 y’ + 4 y = 0 ans. Y = C1 e5x + C2e-6x 4. y”” + y” = 0 , y(0) = y’(0) ans y = ex( ( C1 + C2x) cos x + ( = y”(0), y’”(0) = 1 C3 + C4) sinx ) ans y = x – sin x 5 . y’” + 6y” + 12 y’ + 8y = 0 y(0) = y’(0) , y”(0) = 2 ans y = x2 e-2 Terima kasih V-9