MEC702_Wk_5_Lect_2 - MEC702 Wk 5 Lect 2 Existence and...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MEC702 Wk 5 Lect 2 Existence and Uniqueness of Solutions Theorem. (Existence of a General Solution) If p(x) and q(x) are continuous on an open interval I , then y 00 + p(x)y 0 + q(x)y = 0 has a general solution on I . Theorem. (A General Solution encompasses all solutions) If y 00 + p(x)y 0 + q(x)y = 0 has continuous coecients p(x) and q(x) on some open interval I , then every solution y = V (x) of the ODE on I is of the form V (x) = c1 y1 + c2 y2 where y1 and y2 is any basis of solutions of the ODE on I and c1 , c2 are suitable constants. 1 Non-Homogeneous Linear ODEs We now advance from homogeneous to nonhomogeneous linear ODEs. Consider the second-order nonhomogeneous linear ODE y 00 + p(x)y 0 + q(x)y = r(x) where r(x) 6= 0. Denition. (General Solution, Particular Solution) A general solution of the nonhomogeneous ODE on an open interval I is a solution of the form y(x) = yh (x) + yp (x) ; where yh (x) = c1 y1 + c2 y2 is a general solution of the homogeneous ODE on I and yp (x) is any solution of nonhomogeneous ODE on I containing no arbitrary constants. A particular solution of the nonhomogeneous ODE on I is a solution obtained from y(x) by assigning specic values to the arbitrary constants c1 and c2 in yh . Theorem. (Relation of Solutions) 1. The sum y(x) = yh (x) + yp (x) of a solution yp (x) of the nonhomogeneous linear ODE on some open interval I and the solution yh (x) of the homogeneous linear ODE (with r(x) = 0) on I is a solution of the nonhomogeneous linear ODE. 2. The dierence of two solutions of a nonhomogenous linear ODE on I is a solution of the homogeneous case on I. Theorem. (A General Solution of a Nonhomogeneous ODE Includes all Solutions) If the coecients p(x), q(x) and the function r(x) in y 00 + p(x)y 0 + q(x)y = r(x) are continuous on some open interval I , then every solution of the ODE is obtained by assigning suitble values to c1 and c2 in a general solution y(x) = yh (x) + yp (x) of the ODE on I . 2 Method of Undetermined Coecients The theorems above suggests the following: To solve the nonhomogeneous linear ODE y 00 + p(x)y 0 + q(x)y = r(x) or an IVP, we have to solve the homogeneous ODE and nd the any solution yp of the ODE so as to obtain a general solutions. To nd a solution yp , one method to use is the Method of Undetermined Coecients. This method is suitable for linear ODEs with constant coecients a and b y 00 + ay 0 + by = r(x) when r(x) is an exponential function, a power of x, a cosine or sine, or sums or products of such functions. These functions have derivatives similar to r(x) itself. This gives the good idea. We choose a form for yp similar to r(x), but with unknown coecients to be determined by substituting that yp and its derivatives into the ODE. The table and the rules are given below. Choice Rules for the Method of Undetermined Coecients 1. Basic Rule. If r(x) in ODE is one of the functions in the rst column in the Table below, choose yp in the same line and determine its undetermined coecients by substituting yp and its derivatives into the ODE. 2. Modication Rule. If a term in your choice for yp happend to be a solution of the homogeneous ODE corresponding to the nonhomogeneous ODE, multiply this term by x (or x2 if this solution corresponds to a double root of the characteristic equation of the homogeneous ODE.) 3. Sum Rule. If r(x) is a sum of functions in the rst column of the Table below, choose for yp the sum of the functions in the corresponding lines of the second column. Table for the Method of Method of Undetermined Coecients Term in r(x) γx ke kx , (n = 0, 1, . . .) k cos ωx n k sin ωx keαx cos ωx keαx sin ωx Choice for yp (x) Ceγx Kn x + Kn−1 x + · · · + K1 x + K0 n n−1 K cos ωx + M sin ωx eαx (K cos ωx + M sin ωx) 3 Example. Application of the Basic Rule. Solve the IVP y 00 + y = 0.001x2 , y 0 (0) = 1.5 . y(0) = 0, Solution. Step 1. General Solution of the homogeneous ODE. The ODE y 00 + y = 0 has the general solution yh = A cos x + B sin x. Step 2. Solution yp of the nonhomogeneous ODE. We rst try yp = Kx2 . Then yp00 = 2K . By substitution, y 00 + y = 0.001x2 = 2K + Kx2 = 0.001x2 . For this to hold for all x, we see that K = 0, a contradiction. The second line in the Table suggests the choice yp = K2 x2 + K1 x + K0 . Then for the ODE yp00 + yp = 0.001x2 (2K2 ) + (K2 x2 + K1 x + K0 )  K2 x2 + (K1 x) + (2K2 + K0 ) = 0.001x2 = 0.001x2 Equating the coecient of x2 , x and x0 on both sides, we have K2 = 0.001, K1 = 0 and 2K2 + K0 = 0. Hence This gives and K2 = 0.001 K1 = 0 K1 = −0.002. yp = 0.001x2 − 0.002 y = yh + yp = A cos x + B sin x + 0.001x2 − 0.002. Step 3. Solution of the IVP. Applying the initial condition, we have y(0) = 0 = A − 0.002 =⇒ A = 0.002. By dierentiation and from the second initial condition, y 0 = yh0 + yp0 = −A sin x + B cos x + 0.002x and y 0 (0) = B = 1.5. This gives the answer y = 0.002 cos x + 1.5 sin x + 0.001x2 − 0.002. 4 Example. Application of the Modication Rule. Solve the IVP y 00 + 3y 0 + 2.25y = −10e−1.5x , y 0 (0) = 0. y(0) = 1, Solution. Step 1: General solution of the homogneneous ODE. The characteristic equation of the homogeneous ODE is λ2 + 3λ + 2.25 = 0 =⇒ (λ + 1.5)2 = 0 =⇒ λ = −1.5. Hence, the homogeneous ODE has the general solution yh = (c1 + c2 x)e−1.5x . Step 2: Solution yp of the homogeneous ODE. The function e−1.5x on the right would normally require the choice Ce−1.5x . But we see from yh that this function is a solution of the homogeneous ODE, which corresponds to a double root of the characteristic equation. Hence, according to the Modication Rule, we have to multiply our choice function by x2 . That is, we choose yp = Cx2 e−1.5x . Then yp0 = C(2x − 1.5x2 )e−1.5x yp00 = C(2 − 3x − 3x + 2.25x2 )e−1.5x . We substitute these expressions into the given ODE and omit the factor e−1.5x . This yields yp00 + 3yp0 + 2.25yp = −10e−1.5x C(2 − 6x + 2.25x2 ) + 3C(2x − 1.5x2 ) + 2.25Cx2 = −10. 0Cx2 + 0Cx + 2C = −10 C = −5. Hence the general solution of the given ODE is y = yh + yp = (c1 + c2 x)e−1.5x − 5x2 e−1.5x . Step 3: Solution of the IVP: Applying the initial condition y(0) = 1 = c1 . Dierentiating and applying the second initial condition y0 = (c2 − 1.5c1 − 1.5c2 x)e−1.5x − 10xe−1.5x + 7.5x2 e−1.5x y 0 (0) = 0 = c2 − 1.5c1 gives c2 = 1.5. Then we obtain the full solution of the IVP y = (1 + 1.5x)e−1.5x − 5x2 e−1.5x . 5 Example. Application of the Sum Rule. Solve the IVP y 00 + 2y 0 + 0.75y = 2 cos x − 0.25 sin x + 0.09x, y 0 (0) = −0.43. y(0) = 2.78, Solution. Step 1: General solution of the homogeneous ODE. The characteristic equation of the homogeneous ODE is λ2 + 2λ + 0.75 = 0 which gives the general solution x y = c1 e− 2 + c2 e− 3x 2 . Step 2: Particular solution of the nonhomogeneous ODE. We write yp = yp1 + yp2 and following the Table, we have yp1 = K cos x + M sin x yp2 = K1 x + K0 . Dierentiation gives 0 yp1 = −K sin x + M cos x, 00 yp1 = −K cos x − M sin x, 0 yp2 = K1 , 00 yp2 = 0. For yp1 , substituting into the ODE and equating to r(x), 00 0 yp1 + 2yp1 + 0.75yp1 = 2 cos x − 0.25 sin x (−K cos x − M sin x) + 2(−K sin x + M cos x) + 0.75(K cos x + M sin x) = 2 cos x − 0.25 sin x (−K + 2M + 0.75K) cos x + (−M − 2K + 0.75M ) sin x = 2 cos x − 0.25 sin x −K + 2M + 0.75K = 2 −M − 2K + 0.75M = −0.25 For yp2 , substituting into the ODE and equating to r(x), 00 0 yp2 + 2yp2 + 0.75yp2 = 0.09x (0) + 2K1 + 075(K1 x + K0 ) = 0.09x 6 K0 = −0.32 K1 = 0.12. K = 0 M = 1. Hence the general solution of the nonhomogeneous ODE x y = c1 e− 2 + c2 e− 3x 2 + sin x + 0.12x − 0.32. Step 3: Solution of the IVP. Applying the initial conditions y(0) y 0 (0) 2.78 = c1 + c2 − 0.32 1 3 = −0.4 = − c1 − c2 + 1 + 0.12 2 2 = gives c1 = 3.1 and c2 = 0. The solution of the IVP is then x y = 3.1e− 2 + sin x + 0.12x − 0.32. 7 Solution by Variation of Parameters We continue our discussion of nonhomogeneous linear ODE y 00 + p(x)y 0 + q(x)y = r(x). We have already used the Method of Undetermined Coecients to nd yp when r(x) is not too complicated. However, since this method is restricted to functions r(x) whose derivatives are of a form similar to r(x) itself (powers, exponential functions, etc), it is desirable to have a method valid for general ODEs which we will now see. It is called the Method of Variation of Parameters. Langrage's method gives a particular solution yp of the ODE on some open interval I in the form ˆ yp (x) = −y1 y2 r dx + y2 W ˆ y1 r dx W where y1 , y2 form a basis of solutions of the corresponding homogeneous ODE y 00 + p(x)y 0 + q(x)y = 0 on I and W is the Wronskian of y1 , y2 , y W = 10 y1 y2 . y20 Remark. Caution! The solution formula is obtained under the assumption that the ODE is written in standard form, with y 00 as the rst term. If it starts with f (x)y 00 , divide rst by f (x). Also, the integration in the formula may often cause diculties, and also the wronskian if y1 , y2 has variable coecients. If you have a choice, use the previous method. It is simpler. Example. Solve the nonhomogeneous ODE y 00 + y = sec x = 1 . cos x Solution. A basis of solutions of the homogeneous ODE on any interval is y1 = cos x, y2 = sin x. This gives the Wronskian W (y1 , y2 ) = 1 Using the formula of the variation of parameters method, we have ˆ yp (x) = ˆ y2 r y1 r dx + y2 dx W W ˆ  ˆ  − cos x sin x sec xdx + sin x cos x sec xdx ˆ  ˆ  sin x cos x − cos x dx + sin x dx cos x cos x ˆ  ˆ  − cos x tan xdx + sin x 1dx = − cos x [ln | sec x|] + sin x [x] . = = = −y1 Hence the general solution of the nonhomogeneous ODE is y = yh + yp = c1 cos x + c2 sin x − cos x · ln | sec x| + x sin x. 8...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern