Differential Equations

Transcription

1 Mah 21 (Fall 29) Differenial Equaions Soluion #3 1. Find he paricular soluion of he following differenial equaion by variaion of parameer (a) y + y = csc (b) 2 y + y y = ln, > Soluion: (a) The corresponding homogeneous equaion y +y = has y h () = c 1 sin +c 2 cos as is general soluion, so y 1 = sin, y 1 = cos, y 2 = cos, and y 2 = sin. The Wronskian of y 1 and y 2 is y 1 y 2 W [y 1, y 2 ] = y 1 y 2 = y 1y 2 y 2 y 1 = sin 2 cos 2 = 1 so by variaion of parameer we have cos csc cos v 1 = d = 1 sin d = d sin sin = log(sin ) According, sin csc v 2 = d = 1 y p = sin log(sin ) cos is he desired paricular soluion. (b) The corresponding homogeneous equaion 2 y + y y = is he Cauchy-Euler equaion and {y 1 () =, y 2 () = 1/} is a pair of independen soluion. The Wronskian is y 1 y 2 W [y 1, y 2 ] = y 1 y 2 = ( 1/2 ) (1/) = 2/, > To apply he mehod of variaion parameer, we have o wrie he differenial equaion in he sandard form hen we have v 1 () = v 2 () = y + 1 y 1 2 y = ln, >. y2 () ln W [y 1, y 2 ] d = 1 ln 2 d = 1 2 y1 () ln W [y 1, y 2 ] d = 1 2 ln d = 1 4 ln d ln = 1 (ln )2 4 ln d 2 = ln

2 and v 1 ()y 1 () + v 2 ()y 2 () = 1 4 (ln )2 1 4 ln 1 8. Since 1 8 is a par of he homogeneous soluion, so he required paricular soluion is y p () = 1 4 (ln )2 1 4 ln 2. Le g be a coninue funcion and y be defined by y() = 1 2 sin 2( τ) g(τ)dτ ( ) (a) Apply he Leibniz formula (he generalizaion of he fundamenal heorem of calculus) o show ha he funcion y defined by ( ) saisfies he iniial value problem y + 4y = g(), y() =, y () = (b) Use he variaion of parameers o solve he iniial value problem and show ha he soluion is given by ( ). Soluion : (a) By Leibniz formula we have y () = sin 2( ) d sin 2( τ) g() + g(τ)dτ = 2 d 2 cos 2( τ)g(τ)dτ y d () = cos 2( )g() + cos 2( τ)g(τ)dτ d = g() 2 sin 2( τ)g(τ)dτ = g() 4y() Therefore y saisfies he differenial equaion y +4y = g. I is obvious from he expression of y() and y () ha y() = y () =. (b) Form he homogeneous soluion y h () = c 1 sin 2 + c 2 cos 2 we have y 1 () = sin 2, y 2 () = cos 2, y 1 () = 2 cos 2, y 2 () = 2 sin 2 and he Wronskian W = y 1 y 2 y 2y 1 = 2, hen y 2 (τ)g(τ) v 1 () = W [y 1, y 2 ] dτ = 1 2 v 2 () = y 1 ()g() W [y 1, y 2 ] d = 1 2 cos 2τ g(τ)dτ sin 2τg(τ)dτ 2

3 and he required paricular soluion is y p () = v 1 ()y 1 () + v 2 ()y 2 () = 1 2 = 1 2 (sin 2 cos 2τ cos 2 sin 2τ)g(τ)dτ sin 2( τ)g(τ)dτ Thus he general soluion of he differenial equaion is By Leibniz formula we have y() = c 1 sin 2 + c 2 cos y () = 2c 1 cos 2 2c 2 sin 2 + To deermine he parameers c 1, c 2, we se = hen sin 2( τ)g(τ)dτ cos 2( τ)g(τ)dτ y() = c 2 =, y () = 2c 1 = Therefore he funcion y defined by ( ) is he soluion of he iniial value problem. 3. Find all values of α for which all soluions of approach zero as. Soluion: Subsiuing y = r, we find ha Thus 2 y + αy y = r(r 1) + αr = or r2 + (α 1)r =. r 1 = 1 α + (α 1) and he general soluion is, r 2 = 1 α (α 1) y() = c 1 r 1 + c 2 r 2 In order for soluions o approach zero as i is necessary ha he real pars of r 1 and r 2 be posiive. Suppose ha α > 1, hen (α 1) 2 1 is eiher imaginary or real and less han α 1; hence he real pars of r 1 and r 2 will be negaive. Suppose α = 1, hen r 1, r 2 = ±i 1 and he soluions are oscillaory. Suppose α < 1, hen (α 1) 2 1 is eiher imaginary or real and less han α 1 = 1 α; hence he real pars of r 1 and r 2 will be posiive. Thus, if α < 1 he soluions of he D.E. will approach zero as. 4. Deermine he general soluion of he given differenial equaion ha is valid in any inerval no including he singular poin. 3

4 (a) ( + 1) 2 y + 3( + 1)y +.75y = (b) 2 y + 3y + 5y = (c) 2 y 5y + 9y = Soluion: (a) Assume y = ( + 1) r for + 1 >. Subsiuion of y ino he D. E. yields r(r 1) + 3r = or r2 + 2r = which yields r = 3 2, 1 2. The general soluion of he D.E. is hen y = c 1 x + 1 1/2 + c 2 x + 1 3/2, x 1 (b) If y = r hen r(r 1) + 3r + 5 = or r 2 + 2r + 5 = and r = 1 ± 2i. Thus he general soluion of he D. E. is y() = c 1 1 cos(2 ln ) + c 2 1 sin(2 ln ), (c) Again le y = r o obain r(r 1) 5r + 9 = or (r 3) 2 = and he general soluion is y() = c c 2 3 ln, 5. Find a paricular soluion of y y 6y = e firs by undeermined coefficiens and hen by variaion of parameers. Soluion: The proper paricular soluion should be y p () = Ae, hen subsiuing ino he D.E. we have y p y p 6y p = Ae + Ae 6Ae = 4Ae = e = A = 1 4 We can also guess he paricular soluion by eliminaing he righ hand side; (D 3)(D + 2)(D + 1)y = = y = (c 1 e 3 + c 2 e 2 ) + c 3 e = y h + y p Thus he paricular soluion is y p = c 3 e. Nex, from he homogeneous soluion y h = c 1 y 1 + c 2 y 2 = c 1 e 3 + c 2 e 2 4

5 The Wronskian is W [y 1, y 2 ] = y 1 y 2 y 1 y 2 = e3 ( 2e 2 ) e 2 (3e 3 ) = 5e and v 1 () = v 2 () = Therefore he paricular soluion is y2 ()e W [y 1, y 2 ] d = 1 2 e 4 y1 ()e W [y 1, y 2 ] d = 1 5 e y p () = v 1 ()y 1 () + v 1 ()y 2 () = 1 2 e 4 e e e 2 = 1 4 e 6. Solving he following Cauchy-Euler equaions by using he subsiuion = e x, Y (x) = y() = y(e x ) o change hem o consan coefficien equaion. (a) 2 y + y 9y = (b) 2 y + 3y + y = + 1 (c) 3 y y 5y 15y = 4 Soluion: (a) The chain rule implies he following relaions: and he equaion becomes dy d = dy dx, 2 d2 y d 2 = d2 Y dx 2 dy dx d 2 Y dx 2 dy dx + dy dx 9Y = d2 Y dx 2 9Y = Thus he general soluion of he original equaion is Y (x) = c 1 e 3x + c 2 e 3x = y() = c c 2 3 (b) Same compuaion as (a), he equaion becomes The homogeneous soluion is Y + 2Y + Y = e x + e x Y h (x) = c 1 e x + c 2 xe x By superposiion principle and undeermined coefficiens mehod we compue he paricular soluion Y p = Y 1 + Y 2 separaely; Y 1 + 2Y 1 + Y 1 = e x = Y 1 = 1 4 ex 5

6 Y 2 + 2Y 2 + Y 2 = e x = Y 2 = 1 2 x2 e x Thus he general soluion of he new equaion is Y = c 1 e x + c 2 xe x ex x2 e x Finally, replacing x by ln, we obain a general soluion of he original equaion y() = c c 2 1 ln (ln ) 2 (c) Following he same compuaion as (a) by chain rule we have he 3rd derivaive 3 d3 y = D(D 1)(D 2)Y, d3 D We can ransform he original equaion ino = d dx D(D 1)(D 2)Y = 4D(D 1)Y 5DY 15Y = e 4x From his, by expanding he various operaional producs and hen collecing erms, we find (D 3 + D 2 7D 15)Y = e 4x The auxiliary (characerisic) equaion of his equaion is r 3 + r 2 7r 15 = (r 3)(r 2 + 4r + 5) = Form is roos, r 1 = 3, r 2, r 3 = 2 ± i, we obain he homogeneous soluion Y h (x) = c 1 e 3x + e 2x (c 2 cos x + c 2 sin x) For a paricular soluion we ry Y p (x) = Ae 4x : 64Ae 4x + 16Ae 4x 7(4Ae 4x ) 15(Ae 4x ) = e 4x = A = 1 37 Therefore Y p (x) = 1 37 e4x, and a general soluion is Y (x) = Y h (x) + Y p (x) = c 1 e 3x + e 2x (c 2 cos x + c 3 sin x) e4x Finally, replacing x by ln, we have as a general soluion of he given differenial equaion y() = c 1 e 3 ln + e 2 ln (c 2 cos ln + c 3 sin ln ) + 1 ln e4 37 = c ) (c 2 2 cos(ln ) + c 3 sin(ln ) , 7. Le y 1 be a given nonrivial soluion of he associaed soluion. Find a second linearly independen soluion using reducion of order. 6

7 (a) 2 y 3y + 4y =, y 1 () = 2 (b) y (2 + 1)y + ( + 1)y =, y 1 () = e (c) x 2 y + xy + (x )y =, y 1(x) = x 1/2 sin x, (Bessel equaion) Soluion: (a) To find all soluion from his one, we pu y() = v() 2, y () = v 2 + 2v, y = v 2 + 4v + 2v and subsiue ino he original equaion o obain v 4 + 4v 3 + 2v 2 3v 3 6v 2 + 4v 2 = or v 4 + v 3 = Pu w = v w 4 + w 3 = Assuming for now ha, we ge w + w = The soluion of his separable equaion are w() = C, C an arbirary consan, so we ge v() = w()d = C ln + D Therefore, y() = v() 2 = ( C ln +D ) 2 is he general soluion and y 2 () = 2 ln is he second linearly independen soluion. For convenience, we also apply Theorem 8 (Secion 4.7) o his problem. We rewrie he equaion; y 3 y y = We idenify p() as 3 hen p()d = 3 d = 3 ln and he second independen soluion is e 3 ln y 2 () = y 1 () d = 2 3 y 2 1 d = d = 2 ln (b) We begin by wriing he given equaion in he sandard form y y y = 7

8 Since p() = 2+1, a second linearly independen soluion is given by y 2 = vy 1, where e p()d e 2+ln v() = e 2 d = e 2 d = d = 2 2 This yields y 2 () = v()y 1 () = 2 2 e, so he general soluion is hen y = c 1 e + c 2 2 e (c) Same as (b), we rewrie he original equaion as he sandard form y + 1 ( x y ) 4x 2 y = v(x) = = e p(x)dx y1 2 dx = 1 sin 2 x dx = Thus he second linearly independen soluion is and he general soluion is 1/x (x 1/2 sin x) 2 dx csc 2 xdx = co x = cos x sin x y 2 (x) = v(x)y 1 (x) = x 1/2 cos x y(x) = c 1 x 1/2 cos x + c 2 x 1/2 sin x 8. A weigh of 49g is suspended from a spring of modulus 5 2 g/cm (corresponding o siffness).the coefficien of fricion in he sysem is esimaed o be 1 1 g/(cm/sec). A =, he weigh is pulled down 6 cm from is equilibrium posiion and released from ha poin wih an upward velociy of 2 cm/sec. Find he subsequen displacemen of he weigh as a funcion of ime. When does he weigh pass hrough is equilibrium posiion? Take g = 98cm/sec 2 Soluion: The differenial equaion o be solved is 49 d 2 y 98 d dy 1 d y = or d 2 y d 2 + 2dy d + 5y = The auxiliary (characerisic) equaion os his equaion is r 2 + 2r + 5 =, and is roos are r 1, r 2 = 1 ± 7i. Hence and, differeniaing y = e (c 1 cos 7 + c 2 sin 7) v = dy d = e (c 1 cos 7 + c 2 sin 7) + e ( 7c 1 sin 7 + 7c 2 cos 7) 8

9 Subsiuing he daa y = 6, = ino he equaion for y, we find c 1 = 6 Subsiuing he daa v = 2, = ino he velociy equaion, we find 2 = c 1 + 7c 2, or c 2 = 2 The displacemen of he weigh is hus a damped oscillaion described by he equaion y = e ( 6 cos sin 7), The weigh passes hrough is equilibrium posiion when y =, ha is, when 6 cos sin 7 =, an 7 = 3 Thus = 1 7 Tan 3 + nπ nπ 7 sec 9. The posiion of a cerain spring-mass sysem saisfies he iniial value problem 3 2 y + ky =, y() = 2, y () = v. If he period and ampliude of he resuling moion are observed o be π and 3, respecively, deermine he values of k and v. Soluion: The general soluion is y() = c 1 cos 3 + c 2 sin 3 Subsiuing he iniial condiions ino he equaion yields y() = c 1 = 2, y () = Thus he soluion of he iniial value problem is 3 c 2 = v, c 2 = 3 v The period is The ampliude is A = y() = 2 cos 2π ω = 3 + 2π v2 = 3 v sin = π = k = v2 4 = 3 = v = ±2 5 9

10 1. (Abel formula) If y 1 and y 2 are soluions of he differenial equaion y + p()y + q()y = where p and q are coninuous funcions, hen he Wronskian W [y 1, y 2 ] is given by Abel formula ( W [y 1, y 2 ] = C exp ) p()d (a) Show ha he Wronskian W saisfies he firs order differenial equaion (b) Solve he separable equaion in (a). W + pw = (c) Apply Abel formula o he following differenial equaion Soluion: 2 2 y + 3y y =, >, y 1 = 1/2, y 2 = 1 (a) We sar by noing ha Y 1 and y 2 saisfy y 1 + p()y 1 + q()y 1 = y 2 + p()y 2 + q()y 2 = If we muliply he firs equaion by y 2, muliply he second by y 1, and add he resuling equaions, we obain Nes, we le W = W [y 1, y 2 ] and observe (y 1 y 2 y 1y 2 ) + p()(y 1 y 2 y 1y 2 ) = W = y 1 y 2 y 1y 2 Then we can wrie he previous equaion in he form W + p()w = (b) The firs differenial equaion obained in (a) is separable. ( ) dw = p()d = W = C exp p()d W (c) Direc compuaion shows ha W = 3 2 3/2. On he oher hand, o apply he Abel formula, we mus wrie he differenial equaion in he sandard form so p() = 3/2. Hence W [y 1, y 2 ] = C exp y y y = ( ) 3 2 d = Ce 3 2 ln = C 3/2 This formula gives you he Wronskian of any pair of soluion of he given equaion. For paricular soluions given in his example we mus choose C = 3/2. 1