3. Riley 12.9: The equation sin x dy +2ycos x 1 dx can be reduced to a quadrature by the standard integrating factor,» Z x f(x) exp 2 dt cos t exp (2

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1 PHYS 725 HW #4. Due 15 November Riley 12.3: R dq dt + q C V (t); The solution is obtained with the integrating factor exp (t/rc), giving q(t) e t/rc 1 R dsv (s) e s/rc + q() With q() and V (t) V sin(!t) we thus have»z q(t) e V t/rc R Im t ds e i!s e s/rc» 1 CV Im e i!t e t/rc i!rc +1 CV q 1+(!RC) 2 : 2 4 sin!t tan 1 (!RC) + q!rce t/rc 3 5 : 1+(!RC) 2 We can see this is right by comparing the behavior at small t we should get q(t) ß!V t 2 2. Riley 12.4: The equation 2R : (y x) dy +2x +3y dx is homogeneous of degree 1, so substituting y(x) xv(x) we find (v 1) dv dx +1+(v +1)2 or with v u 1, Z (u 2) du 1 1+u 2 2 ln 1+u 2 Z 2tan 1 u dx x + A: 1

2 3. Riley 12.9: The equation sin x dy +2ycos x 1 dx can be reduced to a quadrature by the standard integrating factor,» Z x f(x) exp 2 dt cos t exp (2 ln(sin x)) sin 2 x ; sin t applying this we have or sin 2 x dy d +2ycos x sin x y sin x 2 sin x dx dx y sin 2 x cos x +1 sin2 x 1 + cos x ; y 1 1+cosx where we have applied the boundary condition y (ß/2) 1 to determine the constant in the solution. 4. Riley 13.6: Use the method of variation of parameters to find the general solutions of (a) d2 y dx 2 y x n Solution: The independent solutions of the homogeneous equation are e x and e x so we let y(x) ff (x) e x + fi (x) e x ; with the subsidiary condition e x ff + e x fi : 2

3 Then differentiating twice and applying the subsidiary condition, we have or Thus d 2 y dx 2 ffex + fie x + e x ff e x fi ; e x ff e x fi x n : y(x) Ae x + Be x + (b) d2 y dx 2 2 dy dx + y 2xex Z x dt t n sinh (x t) : Solution: The independent solutions of the homogeneous equation are e x and xe x so let y (x) ff (x) e x + fi (x) xe x and find fi x 2, ff 2x 3 /3. 5. Riley 13.7: The Green's function is G(x; t) y 2(x)y 1 (t) w(x) (x t)+ y 1(x)y 2 (t) (t x) w(x) where y 2 () 6, y 2 (ß), y 1 (), y ( ß) 6. Since the solutions of the homogeneous equation that satisfy these criteria are y 1 (x) sin (x/2), y 2 (x) cos(x/2), and since the Wronskian is we have w(x) dy 2 dx y 1 dy 1 dx y 2 1 x 2 sin2 1 x 2 2 cos G(x; t) 2 cos (x/2) sin (t/2) (x t) 2 cos (t/2) sin (x/2) (t x) : 6. Riley 14.4 Part (a): zy 2y + zy 3

4 so let y(z) a n z n+ff ; the indicial equation is ff (ff 1) 2ff ; or ff ; 3. We get a 2-term recursion relation (ff + n +2)(ff + n 1) a n+2 + a n : With ff anda 1,the terms are z z2 4 2 z 4 + which we rewrite as z z8 + ::: ( 1) z + 1! 2! z2 3 4! z ! z6 7 8! z8 + ::: (2n 1) ( 1) n+1 z 2n y 2 (z) : (2n)! With ff 3we get, similarly, the terms z z z7 + ::: 3 2 3! 3 n1 z ! 2n ( 1) n+1 z z 7 + ::: 7! (2n +1)! z2n+1 y 1 (z) : 4

5 Part (b): If we expand the sinusoidal functions in power series we get sin z z cos z ( 1) n " z 2n+1 (2n + 1)! ( 1) n+1 (2n +1 1) z 2n+1 (2n + 1)! which is y 1 (z) within the requisite factor 3. z2n+1 (2n)! # ( 1) n+1 2nz 2n+1 (2n + 1)! ; To get the other solution using the Wronskian method we write so that y 2 (z) u (z) y 1 (z) z2 u (z) A [y 1 (z)] : 2 Integrating we find Z z y 2 (z) u (z) y 1 (z) Ay 1 (z) t Z 2 dt [y 1 (t)] z Ay 2 1 (z) or using the hint to perform the integral by parts, y 2 (z) A (z sin z + cos z) : Expanding we recover the series ( 1) z + 1! 2! z2 3 4! z ! z6 7 8! z8 + ::: which we identify with y 2, within amultiplicative factor. Part (c): Calculating the Wronskian we get t 2 dt (sin t t cos t) 2 ; (z sin z +cosz) (sin z z cos z) (z sin z + cos z) (sin z z cos z) (z cos z) (sin z z cos z) (z sin z +cosz) (z sin z) z 2 6 : 5

6 7. Riley 14.5 The equation is y 2zy 2y ; the power series solution about z is y(z) c n z n leading to the recursion relation c n+2 2c n /(n +2); with c 1and c 1 we get y(z) exp z 2 : We can get a second solution using y 1 (z) exp (z 2 ) v(z) which gives v +2zv or v(z) A Z z dt exp t 2 + v() which, with A 1and v() gives y 1 (z) Z z dt exp z 2 t 2 : Butthismust be the power-series solution obtained with c,c 1 1 which is y 1 (z) z 2z 2 n 1 Z (2n + 1)!! z dt exp z 2 t 2 ; where (2n +1)!! df (2n +1) (2n 1) ::: (1). 8. Riley 14.8 The differential equation for the Hermite polynomials is H n 2zH n +2nH n ; 6

7 if we define the generating function G (z; t) df 1 X H n (z) tn n! then the differential equation may bemultiplied by t n /n! and summed to 2 : Since we are given the solution, G (z; t) df H n (z) tn n! exp 2zt t 2 ; we can differentiate with respect to z to get H n(z) tn n! 2t exp 2zt t 2 2 Comparing like powers of t we see that H n (z) tn+1 n! : dh n dz 2nH n 1 : We can also differentiate G with respect to t to get d dt exp 2zt t 2 2(z t) H n (z) tn n! n1 H n (z) ntn 1 n! or t n n [2zH n(z) 2nH n 1(z) H n+1 (z)] : Comparing coefficients of t n we get the desired result, H n+1 2zH n +2nH n 1 : 7

8 9. Riley 14.9 Clearly so that or G (z; t) exp 2zt t 2 exp 2zt t2 z 2 + z 2 exp z 2 exp (z t) 2 exp z 2 H n exp (z t) 2 fi fi fififi t ψ! H n (z) exp z 2 exp z n exp (z t) 2 fi fi fififi t 1. Non-riley problem: The driven, damped oscillator is defined ẍ + fl _x +! 2 x f (t) m Q (t) : Using the operator method (or Laplace transform, or variation of parameters) we find where x (t) dsk (t s) Q (s)+x (t) K (t s) 1 Ω e fl(t s)/2 sin [Ω (t s)] ;! n exp z 2 Ω 2! 2 fl2 4 ; and where x (t)isany solution of the homogeneous equation. Similarly, by direct differentiation or any other method we find _x (t) dsλ(t s) Q (s)+ _x (t) fl 2 x (t) ; 8

9 where Λ(t s) e fl(t s)/2 cos [Ω (t s)] : We now assume Q(t) is a random function with the ensemble averages characteristic of Gaussian white noise: hq(t)i hq(t)q(s)i ff2 ffi (t s) : 2 m Then we can find the expected values and variances of x(t) and _x(t): hx(t)i x (t) ; h _x(t)i _x (t) fl 2 hx(t)i D (x(t) hx(t)i) 2 E! t!1 ff 2 m 2 ff 2 2m 2 Ω 2 fl ds [K (t s)] 2 ψ! fl 2 1 fl 2 +4Ω 2 ff 2 2m 2! 2 fl D (_x(t) h_x(t)i) 2 E ff 2 ψ m 2! t!1 ds [Λ (s)] 2 fl! ff ψ1+ 2 fl2 2m 2 fl 4! + fl 2 2 4! 2fl 2 2 4! 2 ds Λ(s) K (s)+ fl2 4 ff2 2m 2 fl ds [K (s)] 2! Thus, for large t, after the system has settled down, the ensemble average of the (fluctuational) energy of a harmonic oscillator driven by noise is hhi m 2 hd (_x(t) h_x(t)i) 2 E +! 2 D (x(t) hx(t)i) 2Ei ff2 mfl : 9

10 Note that this energy is independent of the oscillator frequency, as long as the oscillator is underdamped. This result is exactly twice the ensemble-averaged kinetic energy which, in the limit that the particle is unbound, is expected to be ff 2 2mfl kt 2 ; that is, the equilibrium thermal energy of an oscillator in a thermal bath at absolute temperature T is kt. 1