Quiz Chapter 08.06: Shooting Method


(All Tests)


(More on Shooting Method)


(More on Ordinary Differential Equations)

Pick the most appropriate answer

1. The exact solution to the boundary value problem

      • \dfrac{d^{2}y}{dx^{2}} = 6x - 0.5x^{2}, \, y(0) = 0, \, y(12) = 0

for y(4) is


2. Given

      • \dfrac{d^{2}y}{dx^{2}} = 6x - 0.5x^{2}, \, y(0) = 0, \, y(12) = 0

the exact value of \dfrac{dy}{dx}(0) is


3. Given

      • \dfrac{d^{2}y}{dx^{2}} = 6x - 0.5x^{2}, \, y(0) = 0, \, y(12) = 0

If one was using Shooting Method with Euler’s Method with a step size of h=4, and assumed value of \dfrac{dy}{dx}(0)=20, then the estimated value of y(12) in the first iteration is most nearly


4. The transverse deflection, u of a cable of length, L, fixed at bothe ends is given as a solution to

      • \dfrac{d^{2}u}{dx^{2}} = \dfrac{Tu}{R} + \dfrac{qx \left( x - L \right)}{2R}


      • T= tension in cable
      • R= flexural stiffness
      • q= distributed transverse load

Given values are L=50", \, T=2000lbs, q=75 lbs/in, R=75 \times 10^{6} lbs⋅in2. The Shooting Method is used with Euler’s Method assuming a step size of h=12.5". Initial slope guesses at x=0 of \dfrac{du}{dx}=0.004 are used in order, and then refined for the next iteration using linear interpolation after the value of u(L) is found. The deflection in inches at the center of the cable found during the second iteration is most nearly


5. The radial displacement, u is a pressurized hollow, thick cylinder (inner radius=5", outer radius=8") is given at different radial locations.

Radius (in.) Radial Displacement (in.)
5.0 0.0038731
5.6 0.0036165
6.2 0.0034222
6.8 0.0032743
7.4 0.0031618
8.0 0.0030769

The maximum normal stress, in psi, on the cylinder is given by

      • \sigma_{max} = 3.2967 \times 10^{6} \left( \dfrac{u(5)}{5} + 0.3 \dfrac{du}{dr}(5) \right)

The maximum stress, in psi, with second order accuracy is

Hint: f(x_{i+1}) = f(x_{i}) + {f}'(x_{i})h + \dfrac{{f}''(x_{i})h^{2}}{2} + \dfrac{{f}'''(x_{i})h^{3}}{6} + \, ... \,,

and f(x_{i+2}) = f(x_{i}) + {f}'(x_{i})(2h) + \dfrac{{f}''(x_{i})(2h)^{2}}{2} + \dfrac{{f}'''(x_{i})(2h)^{3}}{6} + \, ... \,

where x_{i+1} = x_{i} + h


6. For a simply supported beam (at x=0 and x=L) with a uniform load q, the deflection v(x) is described by the boundary value ordinary differential equation as

      • \dfrac{d^{2}v}{dx^{2}} = \dfrac{q(L-x)^{2}}{2EI}, \, 0 \leq x \leq L


      • E= Young’s modulus of elasticity of beam
      • I= second moment of cross-sectional area

This is based on assuming that \dfrac{dv}{dx} is small; if \dfrac{dv}{dx} is not small, then the ordinary differential equation is