ENGI 3424 Engineering Mathematics Problem Set 1 Solutions (Sections 1.1 and 1.2)

Transcription

1 ENGI 344 Engineering Matheatics Proble Set 1 Solutions (Sections 1.1 and 1.) 1. Find the general solution of the ordinary differential equation y 0 This ODE is not linear (due to the product y ). However, it is separable. y y y C y C Redefine the arbitrary constant as C R. The general solution is y R whose graph is a faily of circles, centre at the origin, radius R.. Find the general solution of the ordinary differential equation 3 4y This ODE is not linear but it is separable: 3 y 4 provided y 0 - but y 0 is a solution of the ODE! 1 3 y 4 1 y 0 y 4 C y 4 1 C But the solution y 0 cannot be obtained fro this general solution for any finite choice 1 of C, (although it is a liiting case: li 0 ). C 4 C y 0 is said to be a singular solution of the ODE. Therefore the general solution is 1 y or y0 4 C

2 ENGI 344 Proble Set 1 Solutions Page of Find the coplete solution of the initial value proble y e y, y4 This ODE is not linear (due to the product y ), but it is separable: y y y y e e y e e y e e For the left-side integration, either use the substitution u y y du or f u f u recognize the standard integral f ue du e : y e e C But y(4) = 4 e e C C 0 y e e y But y < 0 when = 4, so we require the negative square root: y 4. A particle of ass is falling vertically in a fluid under the action of gravity. Two B g 0 1 additional forces act on it to retard the otion, a buoyant force of and a drag force directly proportional to the speed, D = v, both directed upwards. The particle is released fro rest at tie t = 0. (a) Find the terinal speed of the particle v in ters of,, and g. (b) Find the speed v(t) of the particle and the distance s(t) that the particle has fallen at all subsequent ties in ters of,,, g and t. (a) The net downward force is F g g v. d dv By Newton s second law, F v ( is constant). dv 1 g v t dv At the terinal speed v v, 0 1 g v 0 v 1 g

3 ENGI 344 Proble Set 1 Solutions Page 3 of 10 Question 4 (continued) dv v (b) The differential equation 1 g is both linear and separable. Solution for v(t) by separation of variables: dv dv v 1 1 g g v 1 g ln v C But v0 0 1 g ln 0 0 C 1 g 1 1 v g g ln v ln ln 1 g 1 g v 1 1 g g e v e 1 g 1 g vt 1 e OR Solution for v(t) by the linear ethod: dv h v 1 g h P e e R P h 1 g e R 1 g e e 1 g 1 h h g v t e e R C e e C C e 1 g 1 g v0 0 C 0 C 1 g Therefore vt 1 e, which is also vt v 1 e.

4 ENGI 344 Proble Set 1 Solutions Page 4 of 10 Question 4(b) (continued) The distance fallen s(t) can be found by siple integration of the speed v(t) : ds 1 g 1 1 g e st t e B 1 g 1 g s0 0 0 B 0 B Therefore 1 g st t e 1 5. As light passes through a pane of glass, its intensity I is attenuated so that the relative di change in intensity is (where is a positive constant and is the depth I within the glass). If 75% of the incident light passes through a pane of glass that is 5 c thic, then how thin ust the pane be for it to allow 99% of the incident light to pass through? This ODE is in a for whose variables have alrea been separated! di ln I C I Let I o be the incident intensity of the light at the surface of the pane (where = 0). ln Io 0 C C ln Io ln I ln Io ln I Io I 4 Io at 5 5 ln ln I at any depth is Therefore the intensity 5 5 I ln ln ln I Io Io

5 ENGI 344 Proble Set 1 Solutions Page 5 of 10 Question 5 (continued) [Thus the intensity decreases by one quarter for every additional depth of 5 c. 9/16 = 56.5% of the incident light will penetrate a 10 c pane. After a etre, less than 1% of the light is still getting through.] Now we require the depth at which 99% of the incident light penetrates I 99% I o ln ln 5 ln ln ln 0.99 The eact answer is 5 c. ln 0.75 Correct to the nearest 0.1, Soe non-separable ODEs can be transfored into separable ODEs by an appropriate change of variables. Use the change of variables v y 1 to convert the ODE y1 e into a separable ODE involving dv, v and. Hence find the solution y of the original ODE as an eplicit function of. y1 e is neither separable nor linear. Differentiating with respect to : dv v y 1 0 y1 dv v dv v e becoes e e which is a separable ODE v v e dv e C v where ln ln e C A A C v e v y A A A y 1 ln A

6 ENGI 344 Proble Set 1 Solutions Page 6 of A etal bar whose teperature is initially 0 C is iersed in boiling water. Deterine the length of tie needed to raise the teperature of the bar to 80 C if at the end of 10 seconds the teperature of the bar is 5 C. Apply a variation of Newton s law of cooling: d o where o 100 C (the abient teperature) and it is nown that Applying the ethod of separation of variables: d d ln 100 C ln C C ln ln 100 ln 80 ln ln ln ln t ln ln t 10 ln ln t 10 ln ln 10 ln ln Therefore, to the nearest second, the tie needed for the bar to reach 80 C is 0 0 and OR 3 35s The ODE is also linear: d h 100 h P e e h e R 100 e 100e h h t e e R C e 100e C 100 C e e 10 e C C 80 t e t 10 / t e e 80 4 t ln e ln t 10ln ln

7 ENGI 344 Proble Set 1 Solutions Page 7 of Find the equation of the faily of curves that is orthogonal to the faily y ce. y ce ce y 1 The orthogonal faily of curves has equations that satisfy, y which is a separable ODE (and note that y = 0 is ipossible unless c = 0). y y y B Let A = B, then the orthogonal faily is or, equivalently, 1 B y. y A A setch of soe representative ebers of both failies of curves supports this conclusion. Even the degenerate curve y0e 0 is orthogonal to all these parabolas!

8 ENGI 344 Proble Set 1 Solutions Page 8 of A potential function V(, y) has equipotential curves that are the faily of all circles that have their centres on the -ais and that touch the y-ais at the origin. The Cartesian equations of the equipotentials are y a, where a is the coordinate of the centre of the circle (and a is the radius of the circle). (a) Setch three representative ebers of the faily of equipotential curves. Four ebers are shown here. All ebers of this faily touch the y-ais at the origin, with the other -ais intercept at the point a, 0. a < 0 places the circle to the left of the y-ais. a = 0 is the single point at the origin (a degenerate circle of zero radius). As a the circle flattens out into a line the y-ais. Show that the corresponding faily of lines of force is another faily of circles that have their centres on the y-ais and that touch the -ais at the origin, as follows: (b) Eliinate the paraeter a to show that all of the equipotentials satisfy the differential y equation. y Differentiating the equation of the equipotentials with respect to iplicitly, y a y a y a Equating the two fors for a : y y y y y y (ecept at the origin y a 0 where is undefined).

9 ENGI 344 Proble Set 1 Solutions Page 9 of 10 Question 9 (continued) (c) Write down the differential equation satisfied by the orthogonal faily of curves. Taing the negative reciprocal, the orthogonal faily of curves ust be the general solution of y y (d) Show that the differential equation in part (c) is equivalent to d y 0 y Starting fro the answer and using the product and chain rules of differentiation: d d 1 y y y 0 y y y 1 y y 0 y y Provided y 0, we obtain y y, y which is the ODE in part (c) above. y Therefore the ODE (which is neither separable nor linear) is equivalent to y d the eact for y y 0. (e) Show that the general solution to the ODE of part (d) can be written as where c is an arbitrary constant of integration. cy y 0 d y 0 y B y By y y Redefine the arbitrary constant of integration as B c and the general solution becoes cy y 0.

10 ENGI 344 Proble Set 1 Solutions Page 10 of 10 Question 9 (continued) (f) Add one eber of the faily of lines of force to your setch in part (a). The circle 0 cy y y c c has radius c and centre (0, c). Again four ebers of the faily are plotted here This plot was created using nothing ore sophisticated than Microsoft Paint! Upon close eaination, one can chec that every pin [thic] circle intersects every blue [thin] circle at right angles; once at the origin and once elsewhere. If the blue circles are equipotentials then the pin circles are the corresponding lines of force and vice versa. This question odels the equipotential curves and lines of force for a dipole. Bac to the inde of solutions