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MATH 1014.03MW Exam 4 Monday, March, 27 2000

NAME:

Student Number:tex2html_wrap_inline240 No.tex2html_wrap_inline242Marks tex2html_wrap_inline244


Instructions: This exam contains 4 questions and has a total of 100 marks. Show all of your work. Time: 35 minutes.

tex2html_wrap_inline248 (20 Marks) Solve the following equation


eqnarray31

Solution (Method 1) Rewrite the equation as
displaymath252

Integrating the above equation from 0 to t, we have
displaymath258

and
displaymath260

This implies tex2html_wrap_inline262.

(Method 2) The general solution of the equation is
displaymath266

Since s(0)=1000, we have C=1000. Hence, the required solution is
displaymath270

tex2html_wrap_inline272(20 Marks)Find the general solution of the equation
displaymath274

Solution Rewrite the equation as
displaymath276

The integrating factor
displaymath278

The general solution
displaymath280

tex2html_wrap_inline282(25 Marks)Find the general solution of the following equation.
displaymath284

Solution (1) Find the general solution of y''+6y'+9y=0.

The auxiliary equation is tex2html_wrap_inline290. Solving the equation, we have r=-3. The general solution
displaymath294

(2) [Note that tex2html_wrap_inline298 and tex2html_wrap_inline300 are particular solutions of y''+6y+9y=0]

Let tex2html_wrap_inline304. Then
displaymath306


displaymath308


eqnarray94

This implies 2C=1 and C=1/2. Hence,
displaymath314

(3) The required general solution is tex2html_wrap_inline318

tex2html_wrap_inline320(25 Marks)Let tex2html_wrap_inline322 and tex2html_wrap_inline324. Prove that
displaymath326

Solution Taking the partial derivatives of tex2html_wrap_inline322, we obtain
displaymath330

This implies tex2html_wrap_inline332 and tex2html_wrap_inline334. Since


displaymath336

and


displaymath338

we obtain
eqnarray172




Kunquan Lan
Mon Mar 27 10:04:47 EST 2000