2302-practice-final-soln

# Solution first we need to write this in the standard

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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...
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Solution First we need to write this in the standard spring...

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