Math 3200 Exam #3
Briggs' Section - Spring 1998

This take-home exam is given for your benefit, so that you may have the best opportunity to demonstrate your understanding of the material. Please respect the ground rules for the exam: You may use books, notes, or other resources, but you may not talk to anyone else about the exam. Collaboration is extremely easy to detect and if I suspect that it has occurred, the burden is on you to prove that it did not happen (contrary to the principles of American law!).

  1. Find the general solution of three of the following ODEs.
    1. tex2html_wrap_inline257 .
    2. tex2html_wrap_inline259 .
    3. y'''(t)-y'(t)=1+5t.
    4. tex2html_wrap_inline263 .
  2. Consider the integral equation

    displaymath201

    Use the convolution theorem to find the Laplace transform of the solution.

  3. Consider the ODE tex2html_wrap_inline265 .
    1. Define a new independent variable t with the relation tex2html_wrap_inline269 and transform this ODE to a constant coefficient ODE.
    2. Find the general solution of the new ODE.
    3. Give the general solution of the original ODE.
  4. Consider the oscillator equation tex2html_wrap_inline271
    1. Briefly (in 2-3 words) describe the physical meaning of tex2html_wrap_inline273 when this ODE models an oscillator.
    2. Let b=0, tex2html_wrap_inline277 , tex2html_wrap_inline279 , and A=6. Use Laplace transforms to solve the ODE subject to the initial conditions y(0)=1 and y'(0)=0.
    3. Express the solution as the product of two trigonometric functions with two different periods. Sketch the solution showing clearly the two periods in the problem (a computer graph is not needed). Briefly interpret the solution.
  5. Find the inverse Laplace transforms of the following functions.

    displaymath202

Math 3200 Exam #3 Solutions
Briggs' Section - Spring 1998

  1. (6 points each) YOU WERE ASKED TO SOLVE ONLY THREE OF THE FOUR ODES! a. The first ODE is a Bernoulli equation. The change of dependent variable tex2html_wrap_inline287 and the substitution tex2html_wrap_inline289 transforms the original ODE the the linear ODE

    displaymath203

    Using the integrating factor tex2html_wrap_inline291 , the general solution is found to be

    displaymath204

    Transforming back to the original variable y(t), we have the general solution

    displaymath205

    b. One number was typed in wrong, so the algebra on this one was worse than expected (though not prohibitive). We have to find the homogeneous solution and the particular solution. For the homogeneous solution, the characteristic polynomial is tex2html_wrap_inline297 which has roots tex2html_wrap_inline299 . So the general solution of the homogeneous problem is

    displaymath206

    For the particular solution, we use a trial solution tex2html_wrap_inline301 . Substituting and matching coeffcients of like powers of t, we find the undetermined coefficients are tex2html_wrap_inline305 , and tex2html_wrap_inline307 . So the general solution of the ODE is

    displaymath207

    c. This is a third order ODE. For the homogeneous solution, the characteristic polynomial is tex2html_wrap_inline309 which has roots tex2html_wrap_inline311 . So the general solution of the homogeneous problem is

    displaymath208

    For the particular solution, note that a homogeneous solution appears on the right side of the ODE, so we use a trial solution of the form tex2html_wrap_inline313 . Substituting and matching terms produces the undetermined coeficient values A=-1 and tex2html_wrap_inline317 . The general solution of the ODE is

    displaymath209

    d. For the homogeneous solution, the characteristic polynomial is tex2html_wrap_inline319 which has roots tex2html_wrap_inline321 . So the general solution of the homogeneous problem is

    displaymath210

    For the particular solution, note that a homogeneous solution appears on the right side of the ODE, so we use a trial solution of the form tex2html_wrap_inline323 . Substitution and matching of terms gives the coefficient values A=0 and B=2. The general solution of the ODE is

    displaymath211

  2. (5 points) Taking the Laplace transform of both sides of the integral equation, we have

    displaymath212

    Noting that the integral is a convolution of tex2html_wrap_inline329 and y(t), we can apply the convolution theorem and simplify:

    displaymath213

    A little algebra gives us the transform of the solution as

    displaymath214

    YOU WERE ASKED TO FIND ONLY THE TRANSFORM OF THE SOLUTION.

  3. (5 points) The problem was to solve the ODE by a change of independent variable, not by a trial solution. We let the new variable t be given by tex2html_wrap_inline269 or tex2html_wrap_inline337 . There are several equivalent ways to do the transformations, but they all rely on the the chain rule in the form

    displaymath215

    Using this fact for the first derivatives, we have

    displaymath216

    Differentiating both sides of this expression with respect to x, we have

    displaymath217

    The product rule gives us

    displaymath218

    Multiplying through by x gives us

    displaymath219

    Now it's a matter of substituting into the ODE and replacing all derivatives with respect to x. We end up with the new constant coeficient ODE y''(t) - 4y(t) = 0, which has the general solution tex2html_wrap_inline347 . Now letting tex2html_wrap_inline337 , we have the general solution tex2html_wrap_inline351 .

  4. (7 points) The parameter b measures the strength of friction in the system, tex2html_wrap_inline355 is related to the restoring force and is the natural frequency of oscillation of the system, A is the amplitude of the external forcing, and tex2html_wrap_inline359 is the frequency of the external forcing. With the given values, the ODE becomes tex2html_wrap_inline361 , subject to the initial conditions y(0)=1 and y'(0)=0. The request was to use Laplace transforms, not undetermined coefficients, to solve the initial value problem. Taking the Laplace transform of both sides, using the initial conditions, and using the cosine transform on the right side, we have

    displaymath220

    A little algebra gives us

    displaymath221

    The partial fraction representation must have the form

    displaymath222

    Solving for the coefficients is not bad! You should quickly get B=D=0 and tex2html_wrap_inline369 . Thus

    displaymath223

    The solution of the initial value problem is

    displaymath224

    The next question was to express this solution as a product of trigonometric functions to facilitate graphing and visualization. As done in class and on page 262 of the text, we can write tex2html_wrap_inline371 . The graph of this solution looks like a low-frequency envelope wave, tex2html_wrap_inline373 , with a period of tex2html_wrap_inline375 , enclosing a high frequency wave, tex2html_wrap_inline377 , with a period of tex2html_wrap_inline379 .

  5. (5 points each) a. We write

    displaymath225

    We can now use the shift theorem and the inverse tranforms of the sine and cosine to find that

    displaymath226

    b. There are several ways to invert this function. One is to use the property for the nth derivative of a Laplace tranform. Perhaps the easiest way is to use the shift property and recall that tex2html_wrap_inline383 . We then have

    displaymath227

    c. This inverse required looking no further than the front of your book where you will find several transform pairs that involve fractional powers of s. In particular, tex2html_wrap_inline387 or tex2html_wrap_inline389 . Combined with the shift property, this gives us

    displaymath228


Bill Briggs
Sun May 17 13:48:52 MDT 1998