MATH 24 EXAM 3 SOLUTIONS

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1 MATH 4 EXAM 3 S Consider the equation y + ω y = cosω t (a) Find the general solution of the homogeneous equation (b) Find the particular solution of the non-homogeneous equation using the method of Undetermined Coefficients assuming that ω ω (c) What is the suitable guess for the particular solution if ω = ω? (d) Suppose ω > ω, but as time evolves, ω is getting closer and closer to ω What happens to the qualitative behavior of the solution in this process? Is this change of behavior consistent with your answer to part (c)? (a) Making the guess y = e rt gives r + ω = and r, = ±iω Hence y h = c cos ω t + c sin ω t (b) Since ω ω, this is a regular case for UC Hence the guess is y p = A cos ω t + B sin ω t Substituting this guess in the equation gives ( ω + ω )(A cos ω t + B sin ω t) = cosω t The solution is B = and A =, so y ω ω p = cos ω t ω ω (c) If ω = ω this is an exceptional case for UC Therefore, the suitable guess is multiplied by t as y p = At cos ω t + Bt sin ω t (d) As ω approaches ω, the y p in part (b) has oscillations that grow without bound When ω = ω, ie, part (c), this is the case of resonance, where the additional factor t corresponds to the unbounded growth of the amplitude Hence, the unbounded growth of A(ω ) is the precursor of resonance Note: a more complete explanation of the limit as ω approaches ω requires incorporating the initial data with the general solution (see lecture on April )

2 Consider the equation y y + y = et t (a) Find the general solution of the homogeneous equation (b) Find the Wronskian of the fundamental solutions of the homogeneous equation (c) Find the particular solution of the non-homogeneous equation using the method of Variation of Parameters (d) Convert this equation to a system of first-order equations and write it using vector-matrix notation (a) Making the guess y = e rt gives r r+ = and r, = Hence y h = c e t +c te t (b) Setting y = e t and y = te t gives W = y y y y = e t (c) Using VOP, y p = v y + v y, f(t) = e t /t, and v v = y f W = tet e t /t e t =, = y f W = et e t /t e t = t Therefore, v = t,v = lnt, and y p = te t (lnt ) Note that another particular solution is just y p = te t ln t, since the te t part is already contained in the homogeneous solution (d) Setting x = y and x = y gives the x system x = Ax+b, with A = [ e t and b(t) = t [

3 3 Consider the matrix A = (a) Find the eigenvalues (b) Find the eigenvectors (c) What is the dimension of the span of all the eigenvectors? (d) Does the algebraic system Ax = have a unique solution? (a) Since A is upper-triangular, λ = and λ = λ 3 = (b) For λ = the augmented form for (A λ I)v = is, which gives the eigenvector (up to scalar multiplication) v = For λ = λ 3 = the augmented form is, which is a deficient case with only one eigenvector, v = (c) The span of v and v is a -dimensional subspace of R 3, ie, a plane (d) Since λ =, it follows that det(a) = and, therefore, Ax = does not have a unique solution 3

4 4 Determine whether the following statements are TRUE or FALSE Note: you must write the entire word TRUE or FALSE You do not need to show your work for this problem (a) Let y y + y = te t cost Then y p = e t (A cos t + B sin t) is a suitable guess for the particular solution (b) Let y +y = t + Then y p = At +Bt+C is a suitable guess for the particular solution (c) Suppose the eigenvalues of a matrix are {,, } Then the matrix is invertible (d) Let A = Then the sum of the eigenvalues of is (e) The eigenvalues of the fourth-order equation y y = are distinct (ie, different from each other) (a) TRUE This is not an exceptional case, since the characteristic roots are r, = and the corresponding homogeneous solutions are e t and te t, whereas, the forcing function, e t cost, corresponds to r = ±i Therefore, the standard guess will work, which is the given y p If you don t believe it, you can check that the solution is y p = e t ( t cos t + sin t) (b) FALSE This is an exceptional case, since the polynomial on the right-hand side shares the underlying solution with one of the roots, r = Hence, the right guess is y p = At 3 + Bt + Ct (c) FALSE Since λ =, it follows that det(a) = and A is not-invertible (d) TRUE The sum of the eigenvalues is equal to Tr(A) = 3 (e) TRUE Setting y = e rt gives r 4 = Taking the square-root gives r = ± and another square-root yields r, = ±, r 3,4 = ±i Hence, the roots are distinct 4

5 [ 3 5 Consider the system of differential equations x = Ax with A = 5 3 (a) Find the eigenvalues (b) Find the eigenvectors (c) Find the general solution (d) Solve the initial value problem with x() = [ 3 (e) What is the stability structure of the equilibrium solution? (a) Using the formula λ Tr(A)λ + A = gives λ + = Therefore, λ, = ±i (ie, α =, β = ) (b) (A λ I)v = gives [ 3 i 5 3 i, [ which gives one eigenvector as v = and the other one (corresponding [ 3 + i to λ ) is v = v = 3 i (c) Since this is a complex case we decompose the first eigenvector to real and imaginary parts as [ [ [ v = = + i p + iq 3 + i 3 and use the formulae for the two eigensolutions, to obtain [ cos t x (t) = p cos t q sin t =, 3 cos t sin t [ sin t x (t) = q cos t + p sin t = cost + 3 sin t With these, the general solution is given by x(t) = c x + c x (d) Setting t = in the general solution (ie, cost = and sint = ) and equating to the initial values gives [ [ c x() = = 3c + c 3 [ and c = and c = Therefore, x(t) = x = THE END 5 cos t 3 cos t sin t