Math 2233 Homework Set 8

Transcription

1 Math 2233 Homewk Set 8. Determine the lower bound f the radius of convergence of series solutions about each given point xo. (a) y +4y +6xy =,x = Since the coefficient functions 4 6x are perfectly analytic f all x, the differential equation thus possesses no singular points. Thus, every power series solution y(x) = n= an (x xo) n will converges f all x and all xo. In particular, the radius of convergence f solutions about xo = will be infinite. (b) (x )y + xy +6xy =,x =4 Since the coefficient functions x x 6x x are both undefined f x =. Therefe, x = is a singular point f this differential equation. Accding to the theem stated in lecture, if y(x) = n= an (x xo) n is a power series solution, then its radius of convergence will be at least as large as the distance (in the complex plane) from the expansion point xo and the closest singularity of the functions p(x) and q(x). In the case at hand, x = 4 and the closest (in fact, the only) singular point of the coefficient functions p(x) and q(x) isx =. Since 4 =3 we can conclude that the radius of convergence of a series solution of the fm y(x) = n= an(x 4) n will be at least 3. In other wds, the series solution will be valide f all x in the interval, equivalently, f all x such that x 4 < 3 <x<7 (c) ( 4+x 2) y +4xy + y =,x = In this case, the coefficient functions both have singularities when 4x 4+x 2 4+x 2 4+x 2 = x = ±2i

2 2 These two singularies crespond to the points (, ±2) when we represent points in the complex plane as points in the two dimenional plane. Under this representation of the complex plane, the expansion point x = cresponds to the point (, ). Therefe the distances between the expansion point and the singularity are dist(2i, ) = dist( 2i, ) = ( ) 2 +(2 ) 2 =2 ( ) 2 +( 2 ) 2 =2 Hence, the minimal distance is 2, and so the radius of convergence of a power series solution about isatleast2. (d) ( +x 2) y +4xy + y =,x =2 In this case the coefficient functions both have singularities when 4x +x 2 +x 2 +x 2 = x = ±i These two singularies crespond to the points (, ±) when we represent points in the complex plane as points in the two dimenional plane. Under this representation of the complex plane, the expansion point x = 2 cresponds to the point (2, ). Therefe the distances between the expansion point and the singularity are dist(i, 2) = ( 2) 2 +( ) 2 = 5 dist( i, 2) = ( 2) 2 +( ) 2 = 5 Hence, the distance between the expansion point and the closest singularity is 5 and so the radius of convergence of a power series solution about the point xo = 2 will be at least Determine the singular points of the following differential equations and state whether they are regular irregular singular points. (a) xy +( x)y + xy = In this case, the coefficient functions are x Since p(x) is undefined f x =, is a singular point. Since the limits lim x (x ) lim( x) = x x lim x (x )2 lim x3 = x both exist, x = is a regular singular point. Alternatively, one could say that because the degree of the singularity of the function p(x) atthepoint x = is less than equal to and the degree of the singularity of the function q(x) is less than equal to 2, we have regular singular point at x =. (b) x 2 ( x) 2 y +2xy +4y = In this case, the coefficient functions are 2 x( x) 2 4 x 2 ( x) 2

3 3 The function p(x) evidently has a singularity of degree at x = and a singularity of degree 2 at x =. The function q(x) has singularities of degree 2 at x = and x =. In der to be a regularity singular point the degree of the singularity of p(x) must not exceed and the degree of the singularity of q(x) must not exceed 2. Therefe, x = is a regular singular point and x = is an irregular singular point.. (c) ( x 2 ) 2 y + x( x)y +(+x)y = In this case,the coefficient functions are x( x) ( x 2 ) 2 = x( x) ( x) 2 ( + x) 2 = x ( x)( + x) 2 4 x 2 ( x) 2 = +x ( x) 2 ( + x) 2 = ( x) 2 ( + x) The function p(x) evidently has a singularity of degree at x = and a singularity of degree 2 at x =. The function q(x) has a singularity of degree at x = and a singularity of degree 2at x =. In der to be a regularity singular point the degree of the singularity of p(x) must not exceed and the degree of the singularity of q(x) must not exceed 2. Therefe, x =isa regular singular point and x = is an irregular singular point. 3. Compute the Laplace transfm of the following functions. (a) f(t) =t Let f(t) = t. Integrating by parts, with L[f] = te t dt we get u = t du = dt te t dt = vdu = uv = (t) ( s e st dv = e st dt v = s e st vdu ) ( ) s e st dt = s 2 e st = s 2 (b) f(t) =t n Now let f(t) =t n. Integrating by parts, with L[f] = t n e t dt u = t n du = nt n dt dv = e st dt v = s e st

4 4 we get t n e t dt = vdu = uv = (t n ) ( s e st = + n L[ n ] s = n t s = n s = n s ) vdu t n e st dt L [ t n 2] n(n )(n 2) s 3 L [ t n 3] ( s e st)(nt n dt) = =. n(n 2) (2) L [t] n! s n+ s n 4. Use the Laplace transfm to solve the given initial value problem. () y y 6y = ; y() =, y () = Taking the Laplace transfm of both sides of the differential equation yields = L[y ] L[y ] L[6y] = ( s 2 L[y] sy() y () ) (sl[y] y()) 6L[y] = s 2 L[y] s() ( ) sl[y]+() 6L[y] = ( s 2 s 6 ) L[y] s +2 L[y] = s 2 s 2 s 6 = s 2 (s +2)(s 3) the differential equation f y becomes an algebraic equation f L[y]. To undo this Laplace transfm we first carry out a partial fractions expansion of the right hand side of the equation f L[y]. s 2 (s +2)(s 3) = A s +2 + B s 2=A(s 3) + B(s +2) s 3 This expansion must be valid f all values of s; in particular when s = 2 and when s =3. In the fmer case we have s = 2 4=( 2) 2=A( 2 3) + B( 2+2)= 5A so we must have A = 4. In the latter case, we have 5 s =3 = (3) 2=A(3 3) + B(3+2)=5B

5 5 so B =. We then have 5 s 2 L[y] = (s +2)(s 3) = 4 5 s s 3 = 4 ]+ ] 5 L[e 2x 5 L[e3x = L [ 4 + ] 5 e 2x 5 e3x Hence, (taking inverse Laplace transfm of both sides) y = 4 5 e 2x + 5 e3x 5. Use the Laplace transfm to solve the given initial value problem. y 2y +2y = ; y() =, y () =. Taking the Laplace transfm of both sides of the differential equation yields = s 2 L[y] sy() y () 2(sL[y] y()) + 2L[y] = ( s 2 2s +2 ) L[y] L[y] = s 2 2s +2 = s 2 2s ++ = (s ) 2 + We now consult a table of Laplace transfm and spot the following identity L [ e at sin(bt) ] = (s a) 2 + b 2 which looks just like the right hand side of our expression f L[y] once we thake a = and b =. We conclude L[y] =L[e x sin(x)] y(x) =e x sin(x) b 6. Use the Laplace transfm to solve the given initial value problem. y 2y 2y = ; y() = 2, y () =. Taking the Laplace transfm of the differential equation we get = s 2 L[y] sy() y () 2(sL[y] y()) 2L[y] = ( s 2 2s 2 ) L[y] 2s +2 = ( s 2 2s 2 ) L[y] 2s +2

6 6 Thus, 2s 2 L[y] = s 2 2x 2 s = 2 s 2 2s + 3 = 2 s (s ) 2 3 s 3 = (s ) (s ) 2 3 To undo this Laplace transfm it is helpful to first look at a table of Laplace transfms. From such a table one finds L[e at cosh(bt)] = L[e at sinh(bt)] = s a (s a) 2 b 2 b (s a) 2 b 2 and so taking a = and b = 3wehave L[y] = 2L[e x cosh( 3x)] [ L e x sinh( ] 3x) 3 [ ] = L 2e x cosh( 3x) e x sinh( 3x) 3 so y(x) =2e x cosh( 3x) e x sinh( 3x) 7. Use the Laplace transfm to solve the given initial value problem. y +2y + y =4e t ; y() = 2, y () =. Taking the Laplace transfm of both sides of the differential equation we get s 2 L[y] sy() y () + 2 (sl[y] y()) + L[y] =L[4e t ] (s 2 +2s + ) L[y] 2s + 4= 4 s + (s +) 2 L[y] = s +3s +3 +2s +3= = 2s2 +3s +7 s + s + s + L[y] = 2s2 +3s +7 (s +) 3 We now determine the partial fractions expansion of the right hand side. The general ansatz P(x) (s + a) = A 3 s + + B a (s + a) + C 2 (s + a) 3 and so we will try to find constants A, B, C such that 2s 2 +5s +7 = A (s +) 3 s + + B (s +) + C 2 (s +) 3. is

7 7 Multiplying both sides by (s +) 3 we get 2s 2 +5s +7=A (s +) 2 + B (s +)+C. Plugging in s = wefind 2 5+7=C Plugging in s = yields Plugging in s = yields We now solve C =4. 7=A + B + C = A + B +4 A + B =3. 4 = 4A +2B + C =4A +2B +4 4A +2B = 2A + B =5. A + B = 3 2A + B = 5 f A and B. Subtracting the first equation from the second we obtain A +=2 A =2. Now the first equation yields 2+B =3 B =. Thus, A =2,B =, and C = 4. Applying this partial fractions expansion to the equation f L[y] now yields L[y] = 2s2 +3s +7 = 2 (s +) 3 s + + (s +) (s +) 3 Now from a Table of Laplace transfms we find L [ n! t n e at] = (s + a) n+ Hence so s + = L[e t ] (s +) 2 = L[te t ] (s +) 3 = 2 L[t2 e t ] L[y] = 2L[e t ]+L[te t ]+4 2 L[t2 e t ] = L [ 2e t + te t +2t 2 e t] Taking the inverse Laplace transfm of both sides we finally get y(t) = ( 2t 2 + t +2 ) e t.

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