Chapter 2. First Order Scalar Equations

Transcription

1 Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions. The echniques were developed in he eigheenh and nineeenh cenuries and he equaions include linear equaions, separable equaions, Euler homogeneous equaions, and exac equaions. Soon his way of sudying differenial equaions reached a dead end. Mos of he differenial equaions canno be solved by any of he echniques presened in he firs secions of his chaper. People hen ried somehing differen. Insead of solving he equaions hey ried o show wheher an equaion has soluions or no, and wha properies such soluion may have. This is less informaion han obaining he soluion, bu i is sill valuable informaion. The resuls of hese effors are shown in he las secions of his chaper. We presen heorems describing he exisence and uniqueness of soluions o a wide class of firs order differenial equaions. y y = cos( cos(y π 0 π

2 .. Separable Equaions... Separable Equaions. More ofen han no nonlinear differenial equaions are harder o solve han linear equaions. Separable equaions are an excepion hey can be solved jus by inegraing on boh sides of he differenial equaion. Definiion... A separable differenial equaion for he funcion y is an equaion which can be pu in he form h(y y = g(, where h, g are given funcions. Remark: A separable differenial equaion is h(y y = g(y has he following properies: The lef-hand side depends explicily only on y, so any dependence is hrough y. The righ-hand side depends only on. And he lef-hand side is of he form (somehing on y y. Example..: (a The differenial equaion y = is separable, since i is equivalen o y ( y { g( =, y = h(y = y. (b The differenial equaion y + y cos( = 0 is separable, since i is equivalen o g( = cos(, y y = cos( h(y = y. The funcions g and h are no uniquely defined; anoher choice in his example is: g( = cos(, h(y = y. (c The linear differenial equaion y = a is separable, since i is equivalen o g( = a(, y y = a( h(y = y. (d The equaion y = e y + cos( is no separable. (e The consan coefficien linear differenial equaion y = a 0 y + b 0 is separable, since i is equivalen o (a 0 y + b 0 y = g( =, h(y = (a 0 y + b 0. (f The linear equaion y = a + b(, wih a 0 and b/a nonconsan, is no separable. From he las wo examples above we see ha linear differenial equaions, wih a 0, are separable for b/a consan, and no separable oherwise. Separable differenial equaions are simple o solve. We jus inegrae on boh sides of he equaion. We show his idea in he following example.

3 3 Example..: Find all soluions y o he differenial equaion y y = cos(. Soluion: The differenial equaion above is separable, wih Therefore, i can be inegraed as follows: y y = cos( g( = cos(, h(y = y. y ( y ( d = cos( d + c. The inegral on he righ-hand side can be compued explicily. The inegral on he lef-hand side can be done by subsiuion. The subsiuion is u = y(, du = y ( d du u d = cos( d + c. This noaion makes clear ha u is he new inegaion variable, while y( are he unknown funcion values we look for. However i is common in he lieraure o use he same name for he variable and he unknown funcion. We will follow ha convenion, and we wrie he subsiuion as y = y(, dy = y ( d dy y d = cos( d + c. We hope his is no oo confusing. Inegraing on boh sides above we ge y = sin( + c. So, we ge he implici and explici form of he soluion, y( = sin( + c y( = sin( + c. Remark: Noice he following abou he equaion and is implici soluion: y y = cos( h(y y = g(, h(y =, y g( = cos(, y y = sin( H(y = G(, H(y = y, G( = sin(. Here H is an aniderivaive of h, ha is, H(y = h(y dy. Here G is an aniderivaive of g, ha is, G( = g( d. Theorem.. (Separable Equaions. If h, g are coninuous, wih h 0, hen h(y y = g( (.. has infiniely many soluions y saisfying he algebraic equaion H(y( = G( + c, (.. where c R is arbirary, H and G are aniderivaives of h and g.

4 4 Remark: An aniderivaive of h is H(y = h(y dy, while an aniderivaive of g is he funcion G( = g( d. Proof of Theorem..: Inegrae wih respec o on boh sides in Eq. (.., h(y y = g( h(y ( d = g( d + c, where c is an arbirary consan. Inroduce on he lef-hand side of he second equaion above he subsiuion y = y(, dy = y ( d. The resul of he subsiuion is h(y ( d = h(ydy h(y dy = g( d + c. To inegrae on each side of his equaion means o find a funcion H, primiive of h, and a funcion G, primiive of g. Using his noaion we wrie H(y = h(y dy, G( = g( d. Then he equaion above can be wrien as follows, H(y = G( + c, which implicily defines a funcion y, which depends on. This esablishes he Theorem. Example..3: Find all soluions y o he differenial equaion y = y. (..3 Soluion: We wrie he differenial equaion in (..3 in he form h(y y = g(, ( y y =. In his example he funcions h and g defined in Theorem.. are given by h(y = ( y, g( =. We now inegrae wih respec o on boh sides of he differenial equaion, ( y ( y ( d = d + c, where c is any consan. The inegral on he righ-hand side can be compued explicily. The inegral on he lef-hand side can be done by subsiuion. The subsiuion is ( y = y(, dy = y ( d y ( y ( d = ( y dy. This subsiuion on he lef-hand side inegral above gives, ( y dy = d + c y y3 3 = c. The equaion above defines a funcion y, which depends on. We can wrie i as y( y3 ( = c. We have solved he differenial equaion, since here are no derivaives in he las equaion. When he soluion is given in erms of an algebraic equaion, we say ha he soluion y is given in implici form.

5 5 Definiion..3. A funcion y is a soluion in implici form of he equaion h(y y = g( iff he funcion y is soluion of he algebraic equaion H ( = G( + c, where H and G are any aniderivaives of h and g. In he case ha funcion H is inverible, he soluion y above is given in explici form iff is wrien as y( = H ( G( + c. In he case ha H is no inverible or H is difficul o compue, we leave he soluion y in implici form. We now solve he same example as in Example..3, bu now we jus use he resul of Theorem... Example..4: Use he formula in Theorem.. o find all soluions y o he equaion y = y. (..4 Soluion: Theorem.. ell us how o obain he soluion y. Wriing Eq. (..4 as ( y y =, we see ha he funcions h, g are given by h(y = y, g( =. Their primiive funcions, H and G, respecively, are simple o compue, h(y = y H(y = y y3 3, g( = G( = 3 3. Then, Theorem.. implies ha he soluion y saisfies he algebraic equaion where c R is arbirary. y( y3 ( 3 = c, (..5 Remark: Someimes i is simpler o remember ideas han formulas. So one can solve a separable equaion as we did in Example..3, insead of using he soluion formulas, as in Example..4. (Alhough in he case of separable equaions boh mehods are very close. In he nex Example we show ha an iniial value problem can be solved even when he soluions of he differenial equaion are given in implici form. Example..5: Find he soluion of he iniial value problem y =, y(0 =. (..6 y Soluion: From Example..3 we know ha all soluions o he differenial equaion in (..6 are given by y( y3 ( = c, where c R is arbirary. This consan c is now fixed wih he iniial condiion in Eq. (..6 y(0 y3 (0 3 = c 3 = c c = 3 y( y3 ( 3 =

6 6 So we can rewrie he algebraic equaion defining he soluion funcions y as he (ime dependen roos of a cubic (in y polynomial, y 3 ( 3y( = 0. Example..6: Find he soluion of he iniial value problem y + y cos( = 0, y(0 =. (..7 Soluion: The differenial equaion above can be wrien as y y = cos(. We know, from Example.., ha he soluions of he differenial equaion are y( = The iniial condiion implies ha sin( + c, c R. = y(0 = c =. 0 + c So, he soluion o he IVP is given in explici form by y( = sin( +. Example..7: Follow he proof in Theorem.. o find all soluions y of he equaion y = y 3. Soluion: The differenial equaion above is separable, wih h(y = 4 + y 3, g( = 4 3. Therefore, i can be inegraed as follows: ( 4 + y 3 (4 y = y 3 ( y ( d = (4 3 d + c. Again he subsiuion y = y(, dy = y ( d implies ha (4 + y 3 dy = (4 3 d + c 0. 4y + y4 4 = c 0. Calling c = 4c 0 we obain he following implici form for he soluion, y 4 ( + 6y( = c.

7 7 Example..8: Find he soluion of he iniial value problem below in explici form, Soluion: The differenial equaion above is separable wih Their primiives are respecively given by, y =, y(0 =. (..8 + y h(y = + y, g( =. h(y = + y H(y = y + y, g( = G( =. Therefore, he implici form of all soluions y o he ODE above are given by y( + y ( = + c, wih c R. The iniial condiion in Eq. (..8 fixes he value of consan c, as follows, y(0 + y (0 = 0 + c + = c c = 3. We conclude ha he implici form of he soluion y is given by y( + y ( = + 3, y ( + y( + ( 4 3 = 0. The explici form of he soluion can be obained realizing ha y( is a roo in he quadraic polynomial above. The wo roos of ha polynomial are given by y +- ( = [ ± 4 4( 4 3 ] y +- ( = ± We have obained wo funcions y + and y -. However, we know ha here is only one soluion o he iniial value problem. We can decide which one is he soluion by evaluaing hem a he value = 0 given in he iniial condiion. We obain y + (0 = + 4 =, y - (0 = 4 = 3. Therefore, he soluion is y +, ha is, he explici form of he soluion is y( = Euler Homogeneous Equaions. Someimes a differenial equaion is no separable bu i can be ransformed ino a separable equaion changing he unknown funcion. This is he case for differenial equaions known as Euler homogenous equaions. Definiion..4. An Euler homogeneous differenial equaion has he form ( y ( = F. Remark:

8 8 (a Any funcion F of, y ha depends only on he quoien y/ is scale invarian. This means ha F does no change when we do he ransformaion y cy, c, ( (cy F = F. (c For his reason he differenial equaions above are also called scale invarian equaions. (b Scale invarian funcions are a paricular case of homogeneous funcions of degree n, which are funcions f saisfying f(c, cy = c n f(y,. Scale invarian funcions are he case n = 0. (c An example of an homogeneous funcion is he energy of a hermodynamical sysem, such as a gas in a bole. The energy, E, of a fixed amoun of gas is a funcion of he gas enropy, S, and he gas volume, V. Such energy is an homogeneous funcion of degree one, E(cS, cv = c E(S, V, for all c R. Example..9: Show ha he funcions f and f are homogeneous and find heir degree, f (, y = 4 y + y y 3, f (, y = y + y 3. Soluion: The funcion f is homogeneous of degree 6, since f (c, cy = c 4 4 c y + c c 5 y 5 + c 3 3 c 3 y 3 = c 6 ( 4 y + y y 3 = c 6 f(, y. Noice ha he sum of he powers of and y on every erm is 6. Analogously, funcion f is homogeneous degree 4, since f (c, cy = c c y + c c 3 y 3 = c 4 ( y + y 3 = c 4 f (, y. And he sum of he powers of and y on every erm is 4. Example..0: Show ha he funcions below are scale invarian funcions, f (, y = y, f (, y = 3 + y + y + y y. Soluion: Funcion f is scale invarian since f(c, cy = cy c = y The funcion f is scale invarian as well, since = f(, y. f (c, cy = c3 3 + c cy + c c y + c 3 y 3 c c c y = c3 ( 3 + y + y + y 3 c 3 ( 3 + y = f (, y. More ofen han no, Euler homogeneous differenial equaions come from a differenial equaion N y +M = 0 where boh N and M are homogeneous funcions of he same degree. Theorem..5. If he funcions N, M, of, y, are homogeneous of he same degree, hen he differenial equaion N(, y y ( + M(, y = 0 is Euler homogeneous.

9 9 Proof of Theorem..5: Rewrie he equaion as M(, y The funcion f(, y = N(, y y M(, y ( = N(, y, is scale invarian, because M(c, cy f(c, cy = N(c, cy = M(, y y cn c n = M(, = f(, y, N(, y N(, y where we used ha M and N are homogeneous of he same degree n. We now find a funcion F such ha he differenial equaion can be wrien as y = F. Since M and N are homogeneous degree n, we muliply he differenial equaion by in he form (/ n /(/ n, as follows, y M(, y (/ n (y/ (y/ ( = N(, y (/ n = M((/, = M(, y = F, N((/, (y/ N(, (y/ where This esablishes he Theorem. F M(, (y/ = N(, (y/. Example..: Show ha ( y y y y = 0 is an Euler homogeneous equaion. Soluion: Rewrie he equaion in he sandard form (y 3 y ( y y = y 3 y y = So he funcion f in his case is given by (y 3 y f(, y =. ( y ( y This funcion is scale invarian, since numeraor and denominaor are homogeneous of he same degree, n = in his case, (cy 3c c y c (y 3 y f(c, cy = c = = f(, y. (c cy c( y So, he differenial equaion is Euler homogeneous. We now wrie he equaion in he form y = F (y/. Since he numeraor and denominaor are homogeneous of degree n =, we muliply hem by in he form (/ /(/, ha is (y 3 y y = (/ ( y (/. Disribue he facors (/ in numeraor and denominaor, and we ge ( (y/ 3 (y/ y = y = F, ( (y/.

10 0 where ( (y/ 3 (y/ F =. ( (y/ So, he equaion is Euler homogeneous and i is wrien in he sandard form. Example..: Deermine wheher he equaion ( y 3 y = is Euler homogeneous. Soluion: If we wrie he differenial equaion in he sandard form, y = f(, y, hen we ge f(, y = y 3. Bu f(c, cy = c c 3 f(, y, y3 hence he equaion is no Euler homogeneous...3. Solving Euler Homogeneous Equaions. In mahemaics we ofen ransform an equaion we canno solve in a sraighforward way ino an equaion we do know how o solve. Theorem..6 ransforms an Euler homogeneous equaion ino a separable equaion, which we know how o solve. Theorem..6. The Euler homogeneous equaion y = F for he funcion y deermines a separable equaion for v = y/, given by v ( F (v v =. Remark: The original homogeneous equaion for he funcion y is ransformed ino a separable equaion for he unknown funcion v = y/. One solves for v, in implici or explici form, and hen ransforms back o y = v. Proof of Theorem..6: Inroduce he funcion v = y/ ino he differenial equaion, y = F (v. We sill need o replace y in erms of v. This is done as follows, y( = v( y ( = v( + v (. Inroducing hese expressions ino he differenial equaion for y we ge ( F (v v v + v = F (v v v = ( = F (v v. The equaion on he far righ is separable. This esablishes he Theorem. Example..3: Find all soluions y of he differenial equaion y = + 3y. y Soluion: The equaion is Euler homogeneous, since f(c, cy = c + 3c y (c(cy = c ( + 3y c (y = + 3y y = f(, y.

11 Nex we compue he funcion F. Since he numeraor and denominaor are homogeneous degree we muliply he righ-hand side of he equaion by in he form (/ /(/, ( y = ( + 3y ( + 3 y y = ( y. Now we inroduce he change of funcions v = y/, y = + 3v. v Since y = v, hen y = v + v, which implies v + v = + 3v v = + 3v v v We obained he separable equaion v = ( + v v We rewrie and inegrae i, v + v v = v = + 3v v v. v + v v d = d + c 0. = + v. v The subsiuion u = + v ( implies du = v( v ( d, so du d u = + c 0 ln(u = ln( + c 0 u = e ln(+c0. Bu u = e ln( e c0, so denoing c = e c0, hen u = c. So, we ge + v = c + = c y( = ± c. Example..4: Find all soluions y of he differenial equaion y (y + + (y + =. Soluion: This equaion is Euler homogeneous when wrien in erms of he unknown u( = y( + and he variable. Indeed, u = y, hus we obain y (y + + (y + = u u + u = u = u ( u. + Therefore, we inroduce he new variable v = u/, which saisfies u = v and u = v + v. The differenial equaion for v is v + v = v + v v = v v v d = d + c, wih c R. The subsiuion w = v( implies dw = v d, so w dw = d + c w = ln( + c w = ln( + c. Subsiuing back v, u and y, we obain w = v( = u(/ = [y( + ]/, so y + = ln( + c y( = ln( + c.

12 Noes. This secion corresponds o Boyce-DiPrima [?] Secion.. Zill and Wrigh sudy separable equaions in [?] Secion., and Euler homogeneous equaions in Secion.5. Zill and Wrigh organize he maerial in a nice way, hey presen firs separable equaions, hen linear equaions, and hen hey group Euler homogeneous and Bernoulli equaions in a secion called Soluions by Subsiuion. Once again, a one page descripion is given by Simmons in [?] in Chaper, Secion 7.

13 3..4. Exercises....- Find all soluions y o he ODE y = y. Express he soluions in explici form....- Find every soluion y of he ODE 3 + 4y 3 y + y = 0. Leave he soluion in implici form Find he soluion y o he IVP y = y, y(0 = Find every soluion y of he ODE y + + y = Find every soluion y of he Euler homogeneous equaion y = y Find all soluions y o he ODE y = + y. y..7.- (HW Find he explici soluion o he IVP ( + y y = y, y( = Prove ha if y = f(, y is an Euler homogeneous equaion and y ( is a soluion, hen y( = (/k y (k is also a soluion for every non-zero k R.