Summer 2007 Final Exam Solutions - Diff. Equations & Laplace Transform, Exams of Differential Equations

Download Summer 2007 Final Exam Solutions - Diff. Equations & Laplace Transform and more Exams Differential Equations in PDF only on Docsity! Summer, 2007 – Final Exam Solutions 1. (20 pts) Find the explicit solution to the following initial value problem: dy dx + 1 2x y = √ x, y(0) = 1 Solution: The equation is linear but not separable. We must first find the integrating factor: µ(x) = exp ( ∫ 1 2x dx ) µ(x) = exp ( 1 2 lnx ) µ(x) = √ x Multiplying the ODE by the integrating factor and simplifying we get: √ x dy dx + √ x 2x y = ( √ x)2 d dx [ √ xy] = x d[ √ xy] = x dx ∫ d[ √ xy] = ∫ x dx √ xy = 1 2 x2 + C We now use the initial condition, y(0) = 1, to find C: √ xy = 1 2 x2 + C √ 0(1) = 1 2 (0)2 + C C = 0 The explicit solution is: y(x) = 1 2 x3/2 2. (20 pts) Complete each part below: (a) Find the general solution to the following second order ODE: y′′ − 6y′ + 5y = 0 (b) Write the form of the particular solution to the following nonhomogeneous, second order ODE: y′′ − 6y′ + 5y = x + sin x Do not solve for the coefficients! Solution: (a) The auxiliary equation is r2−5r+6 = 0 and its roots are r = 1, 5. Therefore, the general solution is: y(x) = c1e x + c2e 5x (b) Using the method of undetermined coefficients, the form of the particular solution is: yp(x) = Ax + B + C cos x + D sinx 3. (20 pts) Compute the following expressions: (a) L { sin 2t + e−3tt2 } (b) L −1 { s + 1 s2 − 3s + 2 } Solution: (a) L { sin 2t + e−3tt2 } = 2 s2 + 4 + 2 (s + 3)3 (b) We first use partial fraction decomposition: s + 1 s2 − 3s + 2 = 3 s − 2 + −2 s − 1 Then we get: L −1 { s + 1 s2 − 3s + 2 } = L −1 { 3 s − 2 − 2 s − 1 } = 3e2t − 2et 4. (20 pts) Find all λ > 0 for which the boundary value problem: y′′ + 4λy′ + 5λ2y = 0, y(0) = 0, y (π 2 ) = 0 has nontrivial solutions, y(x), and find these solutions. Solution: The auxiliary equation is r2 + 4λr + 5λ2 = 0. Its roots are obtained using the quadratic equation: r = −4λ ± √ (4λ)2 − 4(1)(5λ2) 2(1) r = −4λ ± √ 16λ2 − 20λ2 2 r = −4λ ± √ −4λ2 2 r = −4λ ± 2λi 2 r = −2λ ± λi The general solution is then: y(x) = e−2λx[c1 cos(λx) + c2 sin(λx)]