College of Engineering and Computer ScienceMechanical Engineering DepartmentMechanical Engineering 501ASeminar in Engineering AnalysisFall 2004 Number: 17472 Instructor: Larry CarettoNovember 2 Homework SolutionsPage 174, problem 9 – Determine the type and stability of the critical point for the system of equations shown below. Then find a general solution. Finally sketch or plot some trajectories in the phase plane. (Show the details of your work.)2122114862 yydtdyyydtdyWriting this as a matrix equation we have a11 = -2, a12 = -6, a21 = -8, and a22 = -4. The trace of thismatrix is -2 – 4 = -6 and the determinant is (-2)(-4) – (-8)(-6) = -40. The negative determinant indicates that it is a saddle point which is unstable .This problem is solved by converting the system into a second order equation. See the next problem for the solution approach using eigenvectors. The general solution can be found by differentiating the equation for dy1/dt and substituting the equation for dy2/dt into the result. 21121212486262 yydtdydtdydtdydtydFrom the original equation for dy1/dt we see that y2 = (-1/6)dy1/dt – y1/3. Using this result in the equation above for d2y1/dt2 gives. 3612448248621111211212ydtdyydtdyyydtdydtydRearranging this result and placing it into the usual form for a second order homogenous equation gives.040611212 ydtdydtydThe characteristic equation for this differential equation is 2 + 6 – 40 = 0, which can be factored into ( + 10)( – 4) = 0. The roots of the latter equation are  = -10 and  = 4, so the solution to the differential equation for y1 is.tteCeCy421011From the equation we previously derived that y2 = (-1/6)dy1/dt – y1/3 we can find the result for y2 as follows.   tttteCeCeCeCydtdyy42101421011123141061361tttteCeCeCeCy4210142101234316431610The solution in the text has a slightly different form for the e-10t term. In the equation for y1 the constant is 3c1 and in the equation for y2 the constant is 4c2. Since the final value of the constant Engineering Building Room 1333 Mail Code Phone: 818.677.6448Email: [email protected] 8348 Fax: 818.677.7062is determined by the initial conditions, what is important in the result is that the constant multiplying the e-10t term in the y2 equation is 4/3 times the constant multiplying the same term in the y1 equation. This result holds for both the solution here and the one in the text.Some sample trajectories are shown in the figure below for different values of C1 and C2.Page 174, problem 13 – Solve y’’ + ay’ = 0 (a is constant) and plot some of the trajectories.We can define y1 = y and y2 y’ to get the following matrix equation.212121112221211121010yyadtdyayaaydtdyyayaydtdyAyyWe can find the matrix eigenvalues by the usual equation that Det(A – I) = 0. 001)(2 aaaDet IASo the eigenvalues are  = 0 and  = -a. The corresponding eigenvectors are found below. The general equation for the eigenvector components is 0001)(22121xaxxxxa0xIA For  = 0, we have x2 = 0 and x1 arbitrary. Setting x1 = 0 gives the  = 0 eigenvector as [1 0]T. For  = -a, the first equation gives x1 = – x2/a but the second equation becomes 0 = 0 which leaves x2 arbitrary. If we pick x2 = a we have x1 = –1 so our second eigenvector is [-1 a]T. Thus our solution to this problem is.Engineering Building Room 1333 Mail Code Phone: 818.677.6448Email: [email protected] 8348 Fax: 818.677.7062ateaCCyy1012121We see that y1 = C1 – C2e-at and y2 = aC2e-at. Substituting C2e-at = y2/a into the equation for y1 gives the trajectory equation that y1 = C1 – y2/a or y2 = aC1 – ay1. For a fixed value of a, this is theequation for a family of straight parallel lines with a negative slope. (Different lines in the family are determined by the value of the constant C1 which becomes the intercept for the trajectories.)Page 183, problem 7 – Determine the location and type of all the critical points of the equation y’’+ y – y2/2 = 0. (Use linearization of the corresponding system. Show the details of your work.)If we define y1 = y and y2 = y’, we can write this second-order equation as the following system of first-order equations.2),(),(211212222111yyyyfdtdyyyyfdtdyCritical points occur when both f1 and f2 are zero. We can have f1 = 0 only if y2 = 0. However, f1 = 0 gives200212),(111211212yyyyyyyfThus, we have two possible critical points: y1 = 0 and y2 = 0 and y1 = 2 and y2 = 0.We can linearize a quadratic y2 term by using a Taylor series about y = y0. If f(y) = y2, then f’(y) = 2y and f’’(y) = 2 and all higher derivatives are zero. At the expansion point, y = y0, the function and derivative values are f(y0) = y02, then f’(y0) = 2y0, f’’(y0) = 2. Using these derivatives in the Taylor series form gives the following finite series for y2.2000020200020220000022)(2)(22)(2)()(2)(''))((')()(yyyyyyyyyyyyyyyfyyyfyyyfyfyyfThe last set of terms, 2y0y – y02 is the linearized form of the quadratic term y2. Applying this to they22 term that appears in the second differential equation gives the following result. 2122220,10,1120,110,112112yyyyyyyyydtdyWe can write our linearized differential equations in a matrix format as follows.20011020,1210,1yyyydtdbAyyWe can find the matrix eigenvalues by the usual equation that Det(A – I) = 0. 10111)(0,10,120,1 yyyDet IASo the eigenvalues will be real if y1,0 is greater than or equal to one. They will be imaginary if y1,0 is less than one. The trace of the A matrix is zero. If we expand the quadratic about the first Engineering Building Room 1333 Mail Code Phone: 818.677.6448Email: [email protected]