ENGI 2422 First Order ODEs - Separable Page 3-01

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1 ENGI 4 First Order ODEs - Separable Page Ordinary Differential Equations Equations involving only one independent variable and one or more dependent variables, together with their derivatives with respect to the independent variable, are ordinary differential equations (ODEs). Similar equations involving derivatives and more than one independent variable are partial differential equations (to be studied in a later term). 3. Classification; Separation of Variables Eample 3.. Unconstrained population growth can be modelled by (current rate of increase) is proportional to (current population level). With (t) number in the population at time t, d dt k This is a first order, first degree ODE that is both linear and separable. The order of an ODE is that of the highest order derivative present. The degree of an ODE is the eponent of the highest order derivative present. An ODE is linear if each derivative that appears is raised to the power and is not multiplied by any other derivative (but possibly by a function of the independent variable), that is, if the ODE is of the form n k d y ak R k k 0 d A first order ODE is separable if it can be re-written in the form f (y) g() d The solution of a separable first order ODE follows from f ( y) g d Solution of Eample 3..: d d kdt k dt kt + C kt C ln kt + C e e e Let A e C, then the general solution is (t) A e kt (A > 0) - unconstrained eponential growth The value of the arbitrary constant A can be found if the value of is known at any one value of t. Often this information is provided as an initial condition: A (0).

2 ENGI 4 First Order ODEs - Separable Page 3-0 Eample 3.. (continued) (t) A e kt (A > 0). Eample 3.. One of the simplest constrained growth (or predator-prey) models assumes that the rate of increase in, (the population as a fraction of the maimum population), is directly proportional to both the current population and the current level of resources (or room to grow). In turn, the level of resources is assumed to be complementary to the population: resources α. This leads to the constrained population growth model d dt k ( ) Classification of this ODE: st order, st degree, non-linear, separable. d k dt ( ) Partial fractions: A B + Using the cover-up rule: A B 0 ( ) + ( )

3 ENGI 4 First Order ODEs - Separable Page 3-03 Eample 3.. (continued) + d k dt ( ) ln + ln kt + C ln kt + C + e kt kt C kt Ae Ae kt ( ) Ae kt + Ae Ae kt () t Ae kt + Ae The graph is the logistic curve, (for constrained population growth). kt

4 ENGI 4 First Order ODEs - Separable Page 3-04 In this course, the only first order ODEs to be considered will have the general form with the following classification: M(, y) d + N(, y) 0 Type Feature Separable M(, y) f () g(y) and N(, y) h() k(y) Reducible to M(t, ty) t n M(, y) separable and N(t, ty) t n N(, y) for the same n Eact M N y Linear M/N P()y R() or N/M Q(y) S(y) Bernoulli M/N P()y R()y n or N/M Q(y) S(y) n [Only the separable, eact and linear types will be eaminable.] Eample 3..3 Terminal Speed A particle falls under gravity from rest through a viscous medium such that the drag force is proportional to the square of the speed. Find the speed v(t) at any time t > 0 and find the terminal speed v. Newton s Second Law: d dv F ( mv) m dt dt dv m mg bv (Net force weight drag force) dt In standard form, ( bv mg) dt + ( m) dv 0 M( t, v) Nt (, v) type separable. f v only const.

5 ENGI 4 First Order ODEs - Separable Page 3-05 Eample 3..3 (continued) m dv mg bv dt v dv b dt mg m b dv b where mg v k m b dt k Partial fractions: A B + v k v+ k v k v+ k A ( k+ k) k B ( k k) k + v k k v k v k bt ln ln k + m + ( ( v k) ( v k) ) C v k kbt ln + C pt + C, v+ k m kb b mg bg where p m m b m v k e v+ k pt + C pt Ae v k v A e pt + k A e pt v ( A e pt ) k ( + A e pt )

6 ENGI 4 First Order ODEs - Separable Page 3-06 Eample 3..3 (continued) General solution: Initial condition: v(0) 0 () v t k pt ( + Ae ) Ae pt ( + A) k 0 A A Complete solution: pt e mg v() t k pt, where k and p + e b bg m Terminal speed v : 0 v lim v() t k k v + 0 mg b Graph of speed against time: [For a 90 kg person in air, b kg m k 30 ms 00 km/h. v(t) is approimately linear at first, but air resistance builds quickly. One gets within 0 km/h of terminal velocity very fast, in just a few seconds.]

7 ENGI 4 First Order ODEs - Eact Page Eact First Order ODEs Method: The solution to the first order ordinary differential equation can be written in the implicit form M(, y) d + N(, y) 0 u(, y) c, (where c is a constant) u u du d + 0. y If M(, y) and N(, y) can be written as the first partial derivatives of some function u with respect to and y respectively, then Clairaut s theorem, u u y y leads to the test for an eact ODE: M N y from which either u M d (and use u y N to find the arbitrary function of integration f (y) ) or u N. (and use u M to find the arbitrary function of integration g() ) Note that these anti-derivatives are partial: In u M d, treat y as though it were constant during the integration. In u N, treat as though it were constant during the integration. After suitable rearrangement, a separable first order ODE is also eact, (but not all eact ODEs are separable).

8 ENGI 4 First Order ODEs - Eact Page 3-08 Eample 3.. Find the general solution of + y e d. Rewrite as ye d + e 0 M N The test for an eact ODE is positive: M y e N ( y ) u M d ye d ye + f where f (y) is an arbitrary function of integration. u e + y f ( y ) But N(, y) e f'(y) 0 f (y) c Therefore u ye + c c The general solution is y A+ e Check: y ( 0 + ) e A+ e y' + y e (y' + y) e

9 ENGI 4 First Order ODEs - Eact Page 3-09 Eample 3.. (alternative method) Find the general solution of + y e d. Rewrite as ye d + e 0 M N The test for an eact ODE is positive: M N e y u N e u N e ye y + g where g() is an arbitrary function of integration. u y e + g ( ) But M y e g'() g + c u ye + c c y A+ e

10 ENGI 4 First Order ODEs - Eact Page 3-0 Eample 3.. Solve y. d d y M N separable. 0 But M N 0 y Therefore the ODE is also eact. u M d ln + f ( y ) But u y 0 + f '(y) N y f '(y) y y f ( y) + c y u(, y) ln + c c + Check: y A ln y 0 + d y y'

11 ENGI 4 First Order ODEs - Eact Page 3- A separable first order ODE is also eact (after suitable rearrangement). f () g(y) d + h() k(y) 0 f k y d h + g y M N 0 M 0 y N Therefore eact! However, the converse is false: an eact first order ODE is not necessarily separable. A single counter-eample is sufficient to establish this. Eample 3..: (y' + y) e is eact, but not separable.

12 ENGI 4 First Order ODEs - Eact Page 3- Eample 3..3 Find the equation of the curve that passes through the point (, ) and whose slope at any point (, y) is y /. Solution by the method of separation of variables: d y This ODE is separable. y d ln y ln C + ( ln y ln A A e y A (family of parabolae). But (, ) is on the curve A () Therefore y C ) Solution starting with the recognition of an eact ODE: d y yd 0 y which is not eact. ( y) ( ) However, d 0 is eact: y 0 y y

13 ENGI 4 First Order ODEs - Eact Page 3-3 Eample 3..3 (continued) u M d d ln + f y u y 0 + f '(y) But N f ( y) y y f (y) ln y + c u(, y) ln ln y + c c ln y ln + C ln (A ) y A (as before). In general, if an eact ODE is seen to be separable, (or separable after suitable rearrangement), then a solution using the method of separation of variables is usually much faster.

14 ENGI 4 First Order ODEs - Eact Page 3-4 Some Practice Questions Eample 3..4 Find the complete solution of 3 y 4 d y 3 0, y when ½. P 3 y 4 and Q 4 3 y 3 are both of the type f () g(y). Therefore the ODE is separable y d 4 3 d 4 3 y y 4 ln y C 3 ln ln y 4 C + ln 3 ln (A 3 ) y 4 A 3 or 3 y 4 A OR P y y 3 Q Therefore the ODE is eact. Eact method: Seek u(, y) such that u u 3 y and y 4 y u 3 y 4 + c c The [implicit] general solution is 3 y 4 A But (0.5, ) lies on the curve 3 4 A A Complete solution: 3 y 4

15 ENGI 4 First Order ODEs - Eact Page 3-5 Eample 3..5 Solve (y + ) d + ( + ) 0 P y + (y + ) and Q + are both of the type f () g(y). Therefore the ODE is separable. d y+ + A ln y+ ln + + c ln + Therefore the general solution is ( + )(y + ) A OR P y Q Therefore the ODE is eact. Eact method: Seek u(, y) such that u u y+ and y + u ( + )(y + ) A If one cannot spot the functional form for u(, y), then u M d y + d y + d y + + f y u y + f '(y) But N + f '(y) f (y) y + c u (y + ) + y + c c Let A c c +, then ( + )(y + ) A

16 ENGI 4 First Order ODEs - Eact Page 3-6 Eample 3..6 Solve ( ) ( ) cos y + ysin y d cos y + ysin y 0 cos ( y) cannot be epressed in the form f () g(y). This ODE is not separable. P cos ( y) + y sin ( y) P y + sin ( y) + sin ( y) y cos ( y) Q cos ( y) y sin ( y) Q + sin ( y) y cos ( y) P y Therefore the ODE is eact. Seek u(, y) such that u u cos y + y sin y and y cos y y sin y u sin ( y) y cos ( y) A

17 ENGI 4 First Order ODEs Integrating Factor Page Integrating Factor Before introducing the general method, let us eamine a specific eample. Eample 3.3. Convert the first order ordinary differential equation y d + 0 into eact form and solve it, (without direct use of the method of separation of variables). [Note, this ODE is different from Eample 3..3.] The ODE is separable and we can therefore rearrange it quickly into an eact form: d + y 0 However, let us seek a more systematic way of converting a non-eact form into an eact form. Let I(, y) be an integrating factor, such that {(the ODE) I(, y)} is eact: y I d + I 0 M N M y I + y I y N I + I We need a simplifying assumption (in order to avoid dealing with partial differential equations). Suppose that I I (), then I y 0 and I I' () M y N I + 0 I + I' ()

18 ENGI 4 First Order ODEs Integrating Factor Page 3-8 Eample 3.3. (continued) di d I di I d ln I ln + C ln (A) I () A Multiplying the original ODE by I (), we obtain the eact ODE (different from the separable-and-eact form) y A d + A 0 A is an arbitrary constant of integration. Any non-zero choice for A allows us to divide the new ODE by that choice, to leave us with the equivalent eact ODE y d + 0 Therefore, upon integrating to find the integrating factor I (), we can safely ignore the arbitrary constant of integration. If we notice that ( y ) y M and ( y ) y N then we can immediately conclude that u y c so that the general solution of the original ODE is c y

19 ENGI 4 First Order ODEs Integrating Factor Page 3-9 Eample 3.3. (continued) If we fail to spot the form for the potential function u, then we are faced with one of the two longer methods: Either ( ) ( y) u M d y d y d y + f y uy y + f y + f y But N f '(y) 0 f (y) c u y + c c y A y A or ( ) u N y + g u y + g y + g But M y g'() 0 g() c u y + c c y A y A The functional form for the integrating factor for an ODE is not unique. In eample 3.3., I(, y) A n+ y n is also an integrating factor, for any value of n and for any non-zero value of A. Proof: ODE y d + 0 becomes n n n n A y d + y M N 0 M n+ n ( n+ ) y y and N n+ n M ( n+ ) y y the transformed ODE is eact.

20 ENGI 4 First Order ODEs Integrating Factor Page 3-0 Eample 3.3. (continued) One can show that the corresponding potential function is ( y) n+ ( ) u, y + c c n n + (, ) ln + ( ) u y y c c n both of which lead to y A y A Checking our solution: + 0 y A y d + y d 0 Dividing by : yd + 0 ( 0) which is the original ODE.

21 ENGI 4 First Order ODEs Integrating Factor Page 3- Integrating Factor General Method: Occasionally it is possible to transform a non-eact first order ODE into eact form. Suppose that P d + Q 0 is not eact, but that IP d + IQ 0 is eact, where I (, y) is an integrating factor. Then, using the product rule, M I P M I P P + I y y y and N I Q N I Q Q + I From the eactness condition M N I I P Q Q P I y y y If we assume that the integrating factor is a function of alone, then di Q 0 I P Q d y di P Q I d Q y P y Q This assumption is valid only if R Q is a function of only. If so, then the integrating factor is I e R ( ) d [Note that the arbitrary constant of integration can be omitted safely.] Then R( ) d u M d e P, y d, etc.

22 ENGI 4 First Order ODEs Integrating Factor Page 3- If we assume that the integrating factor is a function of y alone, then di P Q di Q P 0 P I y I P y Q P This assumption is valid only if R(y) a function of y only. P y If so, then the integrating factor is ( y) R( y ) I e and u R( y ) N e Q (, y), etc. Eample 3.3. Solve the first order ordinary differential equation ( 3 6 ) ( 3 ) y+ y+ y d+ + y P Q 0 The ODE is not separable. P y y Q P y Therefore the ODE is not eact. P y Q 3 + y P y Q 3 + y Q y 3 + R Rd d I e e e [may ignore arbitrary constant of integration] Therefore an eact form of the ODE is 3y+ 6y+ y e d+ 3 + ye 0 M N

23 ENGI 4 First Order ODEs Integrating Factor Page 3-3 Eample 3.3. (continued) M y ( y) e N ( y) e M y Therefore the ODE is, indeed, eact. If one cannot spot the potential function u(, y) (3 y + ½y ) e, then the net fastest path to a solution is ( 3 ) ( ) u N e y + y + g [Note that an anti-differentiation of M with respect to would involve an integration by parts.] ( 3 6 0) u e y + y + y + + g ( But M 3y+ y + 6 ye ) 0 g g c General solution: ( 3 ) e y + y A

24 ENGI 4 First Order ODEs Integrating Factor Page 3-4 Eample A block is sliding down a slope of angle α, as shown, under the influence of its weight, the normal reaction force and the namic friction (which consists of two components, a surface term that is proportional to the normal reaction force and an air resistance term that is proportional to the speed). The block starts sliding from rest at A. Find the distance s(t) travelled down the slope after any time t (until the block reaches B). AP s(t) AB L W mg N W cos α Friction surface + air F μn + kv () v t ds dt Starts at rest from A v(0) s(0) 0 Forces down-slope: dv m W sinα F mg sinα μmg cosα kv dt dv a bv, where a g( sinα μ cosα) and b dt k m Separation of variables method: dv a bv dt ln a bv t + C b

25 ENGI 4 First Order ODEs Integrating Factor Page 3-5 Eample (continued) ln (a bv) bt + C a bv C 3 e bt a v + Ae b But v(0) 0 bt a 0 + b A Therefore a bt () ( e ) v t b OR Integrating factor method: dv dt a bv (bv a) dt + dv 0 This ODE is not eact. Assume that the integrating factor is a function of t only: I I (t). P bv a P v b Q Q t 0 " Py Q" Pv Qt b 0 b Q Q () R t R dt b dt bt bt () e I t

26 ENGI 4 First Order ODEs Integrating Factor Page 3-6 Eample (continued) The ODE becomes the eact form (bv a) e bt dt + e bt dv 0 M (bv a) e bt, N e bt We want a function u(t, v) such that u bt u bt ( bv a) e and e t v One can easily spot that such a function is a bt a u v e A v + Ae b b But v(0) 0 a 0 + A b bt Therefore a bt () ( e ) v t b v ds a dt b bt ( e ) a bt s t + e + b b C But s(0) 0 a b b C Therefore or s bt a s bt + e b ( sin cos ) kt / m ( e ) mg α μ α m t + k k

27 ENGI 4 First Order ODEs Integrating Factor Page 3-7 Eample (continued) The terminal speed v is v a lim v() t t b mg ( sinα μ cosα) k The object will not begin to move unless F < W sin α a > 0 tan α > μ. [The slope has to be steep enough to overcome the force of static friction.] 3 For a 0 kg block sliding down a slope of angle 30, with μ and k, with 4 g 9.8 ms, the terminal speed is just over 4.5 ms and the graphs of speed and position are

28 ENGI 4 First Order ODEs Integrating Factor Page 3-8 Eample Find the solution of that passes through the origin. + y + y' 0 ( + y) d + 0 P Q The ODE is not separable. P y but Q 0. Therefore the ODE is not eact. Try an integrating factor of the form I I (). P y Q Q 0 R d d Rd I e e R The ODE becomes the eact form ( + y) e d + e 0 u u y (, ) ( ) u y + y e + c c y + Ae y + Ae But (0, 0) is on the curve A A The complete solution is therefore y e Check: + y + y' () + ( e ) + (0 + e ) 0

29 ENGI 4 First Order Linear ODEs Page First Order Linear ODEs The general form of a first order linear ordinary differential equation is P y R d + [or, in some cases, d + Q( y) S( y) ] Rearranging the first ODE into standard form, (P() y R()) d + 0 Written in the standard eact form with an integrating factor in place, the first equation becomes I () (P() y R()) d + I () 0 Compare this with the eact ODE du M(, y) d + N(, y) 0 The eactness condition P I di d M y N di I Pd Let h Pd, then ln I () h() and the integrating factor is h( ) I e, where h P d. The ODE becomes the eact form e h (Py R) d + e h 0 M e h (Py R), N e h and h' P. h h h h h u M d e h y R d y h e d e R d u y e e Rd + f y

30 ENGI 4 First Order Linear ODEs Page 3-30 u h e 0 + y f ( y ) But N e h f '(y) 0 h u y e e Rd + c c h h y e e Rd + h C + is d d + C, where h P d Therefore the general solution of P y R y h h( ) e e R A note on the arbitrary constant of integration in h Let m P d + A h + A and B then m h A h m h A h e e e Be e e e e B m m ( C ) y e e Rd + Pd: e A h h h h C e B e Rd C e e Rd B + + B Therefore we may set the arbitrary constant in h including zero, without affecting the general solution of the ODE at all. Pd to any value we wish,

31 ENGI 4 First Order Linear ODEs Page 3-3 Eample 3.4. Solve y' + y 6e Compare with y' + P() y R() The ODE is linear, with P() and R() 6e. h P d d 3 h 3 e e R d e 6e d 6e d 6 3 ( ) h h y e e R d + C e e + C Therefore the general solution is y e + A e Check: y' e A e 3 y + y 4e + Ae e Ae 6e + 0

32 ENGI 4 First Order Linear ODEs Page 3-3 Eample 3.4. : Find the current I (t) flowing through this simple RL circuit at any time t. The electromotive force is sinusoidal: E(t) E o sin ωt The voltage drop across the resistor is RI. The voltage drop across the inductor is di L dt. Therefore the ODE to be solved is di L dt + RI E sin o ωt di R E sin I ωt + - which is linear. dt L L P R h Pdt R Rt dt L L The integrating factor is therefore e h e Rt/L h e Rd t E L e Rt/ L sinωt dt A two-step integration by parts and some algebraic simplification lead to h Eo Rt/ L e Rdt e ( Rsinωt ωlcosωt) R + ( ωl) Details: D I sin ωt e Rt/L + L / ω cos ωt e Rt L R ω + L / sin ωt e INTEGRATE R Rt L

33 ENGI 4 First Order Linear ODEs Page 3-33 Eample 3.4. (continued) h E o L Rt / L L / sin Rt L L e Rdt e ωt ωe cosωt ω e h Rdt L R R R L h E o L Rt/ L L Rt/ L ω e Rdt e sinωt ωe cosωt + R L R R h E L Rt/ L ωl sin cos ωl R R L + R o e R dt e ωt ωt h E Rt/ L R + ( ωl) ( sinω ω cosω ) o e Rdt e R t L t Introducing the phase angle δ, such that ωl R tan δ, leads to and ( ω ) R R + L cosδ ωl R + ωl sinδ Also sin ωt cos δ cos ωt sin δ sin(ωt δ) Therefore h Eo Rt/ L e Rdt e sin ( ωt δ ) R + ωl ht ht The general solution I () t e e R() t dt + C then becomes () I t ( ω δ) ( ωl) E sin t Ae o + R + stea - state Rt/ L transient For most realistic circuits, the current reaches 99% of its stea state value in just a few microseconds.

34 ENGI 4 First Order Linear ODEs Page 3-34 Eample Find the general solution of the ODE y' y sinh P R sinh Therefore the ODE is linear. h Pd Integrating factor is e h e h e Rd e sinh d e e e e d d e + [Note: integration by parts fails for this integral!] 4 Therefore h ( + ) h e y e e Rd C e + + C 4 y + A e + e 4 ( ) Note: this is equivalent to y + A+ cosh + + A sinh 4 ( ) [because e cosh + sinh and e cosh sinh ]

35 ENGI 4 First Order Linear ODEs Page 3-35 Eample Solve the ODE The ODE is not separable. Re-writing the ODE in standard form, y ( ) y e d + e M N M y e y 0 y ( e ) e d y N e y 0 M y Therefore the ODE is eact. y u M d e d y e + f y u y e + y f ( y ) y But N e f '(y) f (y) y + c u e y y A Therefore the general solution is e y y A OR

36 ENGI 4 First Order Linear ODEs Page 3-36 Eample Alternative Method: y e e y d y ( e ) e d y d + e y Compare with d Q y + S( y) Therefore the ODE is linear (for as a function of y). h Q y h e e y h y y e S e e y h h e e S + C y y e y + C e y C

37 ENGI 4 First Order ODEs Page 3-37 Review for the solution of first order first degree ODEs: P d + Q 0. Is the ODE separable? If so, use separation of variables. If not, then. Can the ODE be re-written in the form y' + P() y R()? If so, it is linear. h h h Pd and y e e Rd C Else + 3. Is P y Q? If so, then the ODE is eact. u u Seek u such that P and Q. y The solution is u c. Else Py Q 4. Is a function of only? Q Rd Py Q If so, the integrating factor is I e, where R. Q Then solve the eact ODE I P d + I Q 0. Else, Q Py 5. Is a function of y only? P Ry Q Py If so, the integrating factor is I( y) e, where R( y). P Then solve the eact ODE I P d + I Q 0. A first order ODE that does not fall into one of the five classes above is beyond the scope of this course.

38 ENGI 4 First Order ODEs Reduction of Order Page Reduction of Order Occasionally a second order ordinary differential equation can be reduced to a pair of first order ordinary differential equations. If the ODE is of the form f (y", y', ) 0 (that is, no y term), then the ODE becomes the pair of linked first order ODEs f (p', p, ) 0, p y' If the ODE is of the form g (y", y', y) 0 (that is, no term), then the ODE becomes the pair of linked first order ODEs dp g p, p, y 0, p d where the chain rule Eample 3.5. d y d d d d Find the general solution of the second order ordinary differential equation Let p y' then p' + p 4 3 dp + p d which is linear. 4 P R h P d d ln ln dp d y" + y' 4 3 dp d p dp h e h ( 4 ) e R d d d 5 5

39 ENGI 4 First Order ODEs Reduction of Order Page 3-39 Eample 3.5. (continued) h h p e e Rd + C y + C + C A y + + B 5 Note that the number of arbitrary constants of integration in the general solution matches the order of the ordinary differential equation. Eample 3.5. d y Solve. d d This second order ODE can be solved by either of the two methods of reduction of order. Let p d Method : The ODE becomes dp p p dp d d p + A (provided p 0) Case p 0: p d + A y B ln + A The case p 0 ( y A ) needs to be considered separately. It is a [trivial] solution of the ODE but it is not included in the general solution. Therefore we need to add the singular solution y A to our general solution. General solution: y B ln( + A) or y A.

40 ENGI 4 First Order ODEs Reduction of Order Page 3-40 Eample 3.5. (continued) Method : The ODE becomes dp dp or 0 p p p p The case p 0 leads to a singular solution (y A) as above. Following the other branch: y+ c y ln p y+ c p e ke d which is separable. y y y e k d e k+ c e + A k This leads to the same general solution as before: y B ln + A or y A

41 ENGI 4 First Order ODEs Reduction of Order Page 3-4 Eample Solve y" y y' Let dp p y y p dp p yp dp y or p 0 p 0 y A, which does satisfy the ODE y" y y'. The other branch is separable. dp y p y + A d This is also separable. d y + A We need to quote some standard integrals: d arctan, + d and C C + a a a d arctanh + a a a C

42 ENGI 4 First Order ODEs Reduction of Order Page 3-4 Eample (continued) If A > 0 then y y + B A + B A A A arctan tan ( ) If A 0 then + B y If A < 0 then A A and y y + B A + B A A A arctanh tanh ( ) The solution y A does not fit into any of these categories. Putting all four cases together, the general solution of y" y y' is: y y c or ( ) Atan A + B A> 0 + B A tanh A + B A< 0 ( A 0) ( )

43 ENGI 4 First Order ODEs - Applications Page Applications Orthogonal trajectories A family of curves in ú can be represented by the ODE f (, y) d Any curve, whose intersection with any member of that family occurs at right angles, must satisfy the ODE. d f y (, ) Another family of curves, all of which intersect each member of the first family only at right angles, must also satisfy the ODE. d f, y Eample 3.6. Find the orthogonal trajectories to the family of curves y c y + y' 0 y c y y The orthogonal trajectories must satisfy the ODE y + y y y d d y + C y A [blue: y c ; red y A ]

44 ENGI 4 First Order ODEs - Applications Page 3-44 Eample 3.6. Find the orthogonal trajectories to the equipotentials of the electric field, Q V (, y) c, 4πε r where r + y. r y y a a + +, where Q 4πε c + yy 0 y y Orthogonal trajectories satisfy y d y + y ln y ln + C ln + ln A y A a family of radial lines. Lines of force and equipotential curves are eamples of orthogonal trajectories. Taking the inverse case to Eample 3.6., for a central force law, the lines of force are radial y lines y k. All of these lines satisfy the ODE. The equipotential curves d must then be solutions of the ODE y y d + c d y The general solution can be re-written as + y r, which is clearly the equations of a family of concentric circles.

45 ENGI 4 First Order ODEs - Applications Page 3-45 Other Applications Newton s Law of Cooling If t time, Θ(t) temperature and Θ f ambient [surrounding] temperature, then the rate at which the temperature of an object decreases, when it is immersed in some other substance whose stea temperature is Θ f, is proportional to the temperature difference: Eample dθ k Θ Θ dt ( () t f ) An object is removed from boiling water (at 00 C) and is left in air at room temperature (0 C). After ten minutes, its temperature has fallen to 60 C. Find its temperature Θ(t). d Θ k( Θ () t 0 ) dt which is separable. [Note that the negative sign has been absorbed into the unknown constant k.] dθ Θ 0 kdt ln ( Θ 0) kt + C kt + C kt C kt Θ 0 e e e Ae Θ 0 + kt Ae But Θ(0) A A 80 and Θ(0) e 0k 0k 60 0 e 80 0k t ln ln kt ln 0 ln kt e t/0 Therefore Θ(t) t/0 0( + ( t/0) ) t /0

46 ENGI 4 First Order ODEs - Applications Page 3-46 Eample A right circular cylindrical tank is oriented with its ais of symmetry vertical. Its circular cross-sections have an area A (m ). Water enters the tank at the rate Q (m 3 s ). Water leaves through a hole, of area a, in the base of the tank. The initial height of the water is h o. Find the height h(t) and the stea-state height h. Torricelli: v gh Change of volume: Change IN OUT dv dt Q a gh But V A h dh A Q b h, where b a g dt This ODE is separable. A dh dt Q b h Let u h / du (/) h / dh dh u du A Q b h u bu dh A du A du Q bu b Q bu A Q bu Q A Q du du b Q bu b Q bu t A Q 0 b b [ τ ] u ln ( Q bu) h h t h ho A Q ( t 0) h ho ( ln( Q b h) ln( Q b ho) ) + b b Therefore

47 ENGI 4 First Order ODEs - Applications Page 3-47 Eample (continued) bt Q Q b h + h ho + ln 0, b a g A b Q b h o t can be found as an eplicit function of h, but not vice-versa. t requires Q b h 0 Q Q h h b b Q ga dv Alternative method: In the stea state, flows in and out balance, so that 0. dt dv Setting 0 in the ODE: dt Q Q 0 Q a gh h h a g ga Two additional resources are available on the web site: More Eamples of ODEs (at and Eamples of Partial Fractions (at Also see other eamples in problem sets and in past tests and eaminations. END OF CHAPTER 3