College of Engineering and Computer ScienceMechanical Engineering DepartmentMechanical Engineering 501ASeminar in Engineering AnalysisFall 2004 Number: 17472 Instructor: Larry CarettoSeptember 28 Homework SolutionsPage 39, problem 15 – Solve the following initial value problem: y’ + 4y = 20, with y(0) = 2. Show the details of your work.This is an example of the general first order linear equation: y’ + f(x)y = g(x) with f(x) = 4 and g(x) = 20. The solution to this equation, for the general case can be adapted to the problem considered here as shown below.dxeCeydxxgeCeydxdxdxxfdxxf20)(44)()(Performing the indicated integrations gives our solution with the general constant, C. 54202044444xxxxxCeeCedxeCeyWe select C to meet the initial condition that y(0) = 2. This gives y(0) = 2 = Ce0 + 5 = C + 5, so C = -3 and our solution is y = 5 – 3e-4x.We can verify this solution by substituting it into the differential equation. Since dy/dx = 12e-4x, this substitution gives y’ +4y = 12e-4x +4(5 – 3e-4x) = 20 which matches the right hand side. Also, y(0) = 5 – 3e0 = 2, matching the initial condition.Page 39, problem 21 – Solve the following initial value problem: y’ = 1 + y2, with y(0) = 0. Show the details of your work.This is an example of a separable first order differential equation. Writing y’ as dy/dx and multiplying both sides of the equation by (1 + y2) allows us to perform an indefinite integration on each side.CxyCdxydyydxdy)(tan11122Since tan-1(0) = 0, the initial condition gives C = 0 and our solution is tan-1(y) = x or y = tan(x) .Engineering Building Room 1333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062September 28 homework solutions ME501A, L. S. Caretto, Fall 2004 Page 2Page 58, problem 1 – Show that the initial value problem xy’ = 4y, with y(0) = 1 has no solution. Does this contradict our present existence theorem?The equation xy’ = 4y is another example of a separable first order differential equation. Writing y’ as dy/dx and multiplying both sides of the equation by dx/(xy) allows us to perform an indefinite integration on each side.Cxyxdxydyydxdyx )ln()ln(44We cannot apply the initial condition that y(0) = 1 to the ln(x) term. Thus, we cannot find a solution to this differential equation with the given initial condition. This is consistent with the existence theorem which requires the derivative, f(x,y), in the differential equation dy/dx = f(x,y) tobe bounded in the region of the solution. At the proposed initial condition, dy/dx = 4y/x is indefinite and therefore not bounded. Thus we do not expect a solution to exist for the given differential equation and initial condition.Page 58, problem 4 – Find all initial conditions such that the initial value problem (x2 – 2x)y’ =2(x – 1)y, with y(x0) = y0 has (a) no solution, (b) more than one solution, and (c) precisely one solution.If we rewrite this differential equation in the form used in the existence and uniqueness theorems we have the following resultxxxyfxxyxyxfdxdy2)1(22)1(2),(22We see that the value of f(x,y) is not bounded at x = 0 and x = 2. Thus there are no solutions at this point. The value of f/y is bounded at all points except the same points where there is no solution. Thus, it does not appear that there are any regions where a solution exists and is not unique.We can check this by trying to solve the differential equation, which has a separable solution. Rearranging the original equation gives the following result.CdxxxdxxxxCdxxxxyxxxydy22222)1(2ln2)1(22222We can evaluate the integrals on the left side of the solution for ln y as follows.)2ln(222222xdxxdxxxxxxdxxxdxxxln)2ln(121222Substituting these integrals into the equation for ln y gives the following result.CxxCxxxCdxxxdxxxxy ln)2ln(ln)2ln()2l n(22222ln22September 28 homework solutions ME501A, L. S. Caretto, Fall 2004 Page 3Taking the exponential of both sides of this equation and rearranging givesxCxeeeeeyCxxCxxy)2(ln)2ln(ln)2ln(lnWe can check this solution by differentiating it and substituting it into our original differential equation.)2()1(2)2()1(21)1(2)2(2xxxxCxxCdxdyyxCCxxCdxdyThe final result above matches our differential equation showing that we have the correct solution. Now let’s see what we can do with this solution, y = Cx(x-2). If x = 0 or x = 2, the only solution we have is the trivial one that y = 0. We cannot fit an arbitrary boundary condition y(x0) =y0 at these points. At any other points, the initial condition that y(x0) = y0 gives C = y0/x0/(x0 – 2). So our solution is y = y0x(x – 2)/x0/(x0 – 2).Page 71, problem 17 – Verify that the functions cos 3x and sin 3x form a basis of solutions for the differential equation y’’ + 9y = 0 and solve the initial value problem with y(0) = 4 and y’(0) = -6. Differentiating the proposed solutions two times and substituting the results for y and y’’ into the original differential equation shows that both solutions satisfy the differential equation: -9 sin 3x + 9 sin 3x = 0; -9 cos 3x + 9 cos 3x = 0. Since the sine and cosine are two linearly independent functions, they form a basis for the second-order differential equation.To match the given initial conditions, we use a linear combination of the two solutions to write y = A sin 3x + B cos 3x. The first initial condition gives y(0) = 4 = A sin 0 + B cos 0 = B, so B = 4. Thesecond initial condition gives y’(0) = 3A cos 0 – 3B sin 0 = 3A = -6 so A = 2. Thus the solution is y = 2 sin 3x + 4 cos 3xPage 80, problem 13 – Solve the following initial value problem: 9y’’ + 6y’ + y = 0; y(0) = 4; y’(0) = -13/3. (Show each step.) The solution to the differential equation y’’ + ay’ + by = 0 depends on the roots of the characteristic equation 2 + a + b = 0. For this problem we have to divide the given differential equation by the y’’ coefficient to obtain the following result y’’ + 2y’/3 + y/9 = 0. The solution to thecharacteristic equation with a = 2/3 and b