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### prob set 9

Course: MATH 409, Fall 2011
School: Texas A&M
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Word Count: 881

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A&amp;M Texas University Department of Mathematics Volodymyr Nekrashevych Fall 2011 MATH 308 Homework 9 9.1. Find the fundamental matrix (t) = eAt satisfying (0) = I for the system x= 3 2 2 2 x. 4 2 8 4 x. 9.2. Find the general solution of the system x= 9.3. Find the general solution of the system 100 x = 4 1 0 x. 362 9.4. Find the general solution of the system x= 2 5 1 2 x+ et t . 9.5....

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A&M Texas University Department of Mathematics Volodymyr Nekrashevych Fall 2011 MATH 308 Homework 9 9.1. Find the fundamental matrix (t) = eAt satisfying (0) = I for the system x= 3 2 2 2 x. 4 2 8 4 x. 9.2. Find the general solution of the system x= 9.3. Find the general solution of the system 100 x = 4 1 0 x. 362 9.4. Find the general solution of the system x= 2 5 1 2 x+ et t . 9.5. Solve the previous system in MatLab using the initial condition x(0) = plot the solution (x1 (t), x2 (t)) on the interval [0, 4 ]. 1 1 and 9.1 Let us solve the system at rst. The characteristic polynomial is 2 2, hence the eigenvalues are = 1 and = 2. For = 1, the eigenvectors are solutions of the system 4 2 2 1 1 2 so we can take v = , . The corresponding solution is x(t) = et et . 2et For = 2: 1 2 2 1 so we can take v = 0 0 v= 2 4 0 0 v= e t 2et 1 2 (0) = and (0)1 = 1 3 2e2t e2t 2 1 1 2 = , . The corresponding solution is x(t) = A fundamental matrix is (t) = 1 2 2e2t e2t . . Then , 2 1 . It follows that (t) = et 2e2t 2e2t et 1/3 2/3 2/3 1/3 = 2et 2e2t 3 4e2t et 3 4e2t et 3 2et 2e2t 3 . 9.2 The characteristic polynomial is 2 . So, the only eigenvalue is 0. An 1 eigenvector is v = . The corresponding solution is constant x(t) = v. 2 A solution of 4 2 1 w= 8 4 2 is, for example, w = 1/4 0 . Then, the corresponding solution of the system 1 2 A general solution of the system is of dierential equation is x(t) = t x(0) = c1 1 2 + + c2 t 1 2 1/4 0 + c2 . 1/4 0 . 9.3 The eigenvalues of the matrix are the entries on the diagonal (since it is triangular). 1 The eigenvalue = 1 is repeated. Its eigenvector is found from the system =0 0 4a =0 3a + 6b + c = 0 We see that a = 0, and b, c are any numbers such that 6b + c = 0. We get 0 an eigenvector 1 . All the other eigenvectors are proportional to it. 6 Therefore, we have to solve the system =0 0 4a =1 3a + 6b + c = 6 We can take a = 1/4, b = 0, and c = 6 + 3/4 = 21/4. We get in this way two solutions of the system: 1/4 0 0 0 . x1 (t) = et 1 , x2 (t) = tet 1 + et 21/4 6 6 For the eigenvalue = 2 we solve the a 4a b 3a + 6 b system =0 =0 =0 0 An eigenvector is 0 , and the corresponding solution is 1 0 x3 (t) = 0 . e2t The general solution is then x(t) = c1 x1 (t) + c2 x2 (t) + c3 x3 (t). 9.4. The characteristic of polynomial the homogeneous system is 2 + 1, hence the eigenvalues are = i. Let us nd the eigenvector associated with = i. (2 i)a 5b = 0 a + (2 i)b = 0 A solution is 2+i 1 . Hence a complex solution of the homogeneous system 2 is x1 x2 2+i 1 = eit = (cos t + i sin t) 2+i 1 = 2 cos t sin t + i(2 sin t + cos t) cos t + i sin t 2 cos t sin t cos t = 2 sin t + cos t sin t +i . It follows that the general solution of the homogeneous system is x = c1 2 cos t sin t cos t + c2 2 sin t + cos t sin t , and the fundamental matrix is (t) = 2 cos t sin t cos t 2 sin t + cos t sin t . Its determinant is (2 cos t sin t) sin t (2 sin t + cos t) cos t = sin2 t cos2 t = 1. Hence, 1 (t) = sin t 2 sin t + cos t cos t 2 cos t + sin t . Then the general solution of the equation is given by the indenite integral (one can also take a denite integral and then add the general solution of the homogeneous system to it): (t) et t 1 (s) (t) dt = et sin t + 2t sin t + t cos t et cos t 2t cos t + t sin t (t) The integrals t cos t dt and t cos t dt = dt = et sin t + 2t sin t + t cos t dt et cos t 2t cos t + t sin t dt t sin t dt are computed by parts: t d sin t = t sin t sin t dt = t sin t + cos t + c, and t sin t dt = t d cos t = t cos t + 3 cos t dt = t cos t + sin t + c. The integrals cos tet dt and sin tet dt can be also computed using integration by parts (twice, and then solving a linear equation in terms of the integral), or by computing 1 (1+i)t e +c= 1+i et (cos t + sin t) et (sin t cos t) 1i t (e cos t + iet sin t) + c = +i + c. 2 2 2 (cos t + i sin t)et dt = et+it dt = e(1+i)t dt = Taking the real and imaginary parts, we get et cos t dt = et (cos t + sin t)/2 + c and et sin t dt = et (sin t cos t)/2 + c. Thus, we have that the general solution of the system is (t) et (cos t sin t)/2 2t cos t + 2 sin t + t sin t + cos t + c1 et (cos t + sin t)/2 2t sin t 2 cos t t cos t + sin t + c2 2 cos t sin t 2 sin t + cos t cos t sin t 3t 2 e 5t 1t 2 e 2t + 1 + c1 = et (cos t sin t)/2 + (1 2t) cos t + (2 + t) sin t + c1 et (cos t + sin t)/2 + (1 2t) sin t (2 + t) cos t + c2 2 cos t sin t cos t + c2 2 sin t + cos t sin t . A dierent approach is to use the Laplace transform with initial conditions 1 0 x(0) = and x(0) = , and then take linear combination of the two 0 1 solutions. 4 =
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