This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solution: First we need to write this in the standard spring equation form of my ′′ + by ′ + ky = f ( t ). In this case, m = 2, b = 0, k = 10, and f ( t ) = 0, so we have 2 y ′′ + 10 y = 0 . This has the characteristic polynomial 2 r 2 + 10 = 0, with roots r = ± √ 5 i . Therefore the solution is y = C cos √ 5 t + D sin √ 5 t. The initial conditions can be interrupted as y (0) = 1 and y ′ (0) = 1. From these we find that C = 1 and D = 1 / √ 5 , giving the solution of y = cos √ 5 t sin √ 5 t √ 5 . Now to answer the question, we want to express y in the form A sin( ωt + φ ). We do this using the angle addition formula; we want to find A , φ , and ω so that y = 1 cos √ 5 t + 1 √ 5 sin √ 5 = A sin φ cos ωt + A cos φ sin ωt, so we need to solve the system of equations A 2 = 1 2 + parenleftbigg 1 √ 5 parenrightbigg 2 = 6 5 , tan φ = 1 1 / √ 5 = √ 5 . MAP 2302, Fall 2011 — Final Exam Review Problems 9 This shows that the solution is y ( t ) = radicalbig 6 / 5 sin parenleftBig √ 5 t + arctan( √ 5) parenrightBig . The amplitude is radicalbig 6 / 5 , the period is 2 π / ω = 2 π / √ 5 , and the phase angle is φ = arctan( √ 5). a. Solve the initial value problem y ′ = (1 + y 2 )tan x with initial condition y (0) = √ 3. Solution: This equation is separable, so we have 1 1 + y 2 dy dx = tan x, integraldisplay 1 1 + y 2 dy dx dx = integraldisplay tan xdx, integraldisplay 1 1 + y 2 dy = integraldisplay tan xdx, arctan y = ln(cos x ) + C. We can find C now by plugging in x = 0: arctan √ 3 = ln(cos 0) + C (= C ) . So the final answer is y ( x ) = tan( ln(cos x ) + arctan √ 3) . If we wanted to, we could further simplify this; arctan √ 3 = π / 3 , and tan is an odd function, so tan( x ) = tan x , so y ( x ) = tan(ln(cos x ) π / 3 ) . b. Use the substitution v = y / x to solve the differential equation dy dx = y 2 + 2 xy x 2 . Please remember to express your solution as a function of y . MAP 2302, Fall 2011 — Final Exam Review Problems 10 Solution: We are told to substitute v = y / x , but what do we substitute for dy / dx ? We find this by differentiation: v = y x , y = xv, dy dx = v + x dv dx (by the product rule) . Now we are ready to make the substitution: dy dx = y 2 + 2 xy x 2 , dy dx = parenleftBig y x parenrightBig 2 + 2 y x , v + x dv dx = v 2 + 2 v, x dv dx = v 2 + v, 1 v 2 + v dv dx = 1 x , integraldisplay 1 v 2 + v dv dx dx = integraldisplay 1 x dx, integraldisplay 1 v 2 + v dv = integraldisplay 1 x dx, The integral on the left is a bit tricky. We need to use partial fractions to integrate it: 1 v 2 + v = 1 v ( v + 1) = A v + B v + 1 , so by clearing denominators we see that we need 1 = A ( v + 1) + Bv, which gives us A = 1 and B = 1. Therefore integraldisplay 1 v 2 + v dv = integraldisplay 1 v 1 v + 1 dv = ln v ln( v + 1) = ln parenleftbigg v v + 1 parenrightbigg . MAP 2302, Fall 2011 — Final Exam Review Problems 11 Since the integral of...
View
Full Document
 Fall '08
 TUNCER
 Laplace, lim, Constant of integration, Exam Review Problems

Click to edit the document details