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Unformatted text preview: Section 3.1 19. 21. 30. 31. 51. If y=1+J3€ then yy"+(y')2 = (1+x/x)(—x_3’2/4)+(x‘m/2)2 = —x‘3’2/4 at 0. Linearly independent, because )63 = + x2 Ix! if x > 0, whereas x3 = — x2 x if x < 0. (a) 3’2 =x3 and Y2 =x3l are linearly independent because x3 =Cx3 would
requirethat c = 1 with x = 1, but 6 = —l with x = —1. (b) The fact that W(yl, y2) = O everywhere does not contradict Theorem 3, because
when the given equation is written in the required form y" — (3/x)y' + (3/x2)y = o,
the coefﬁcient functions p(x) = —3/x and q(x) = 3/x2 are not continuous at x = 0. W021, y2) = —2x vanishes at x = 0, whereas if y] and y; were (linearly independent)
solutions of an equation y" + py' + qy = 0 with p and q both continuous on an open
interval I containing x = 0, then Theorem 3 would imply that Wat 0 on I. (a) The substitution v = In x gives '_d_y dlﬂ _ ld_y
dx dvdx xdv Then another differentiation using the chain rule gives y"  dzy _ 1(a)  any.) y " abc2 ' dx dx dx x dv
1 dy 1 d(dy)dv 1 dy 1 dzy
= ‘7'—+*'— — — = "‘7'—+—2‘° :
xdvxdvdvdx xdvxdv Substitution of these expressions for y’ and y” into Eq. (21) in the text then yields
immediately the desired Eq. (22): dzy
dv2 a dy
+ b— —+ = 0.
( a) f Cy (b) If the roots r1 and r2 of the characteristic equation of Eq. (22) are real and
distinct, then a general solution of the original Euler equation is ’i ’2
y(x) = cle"”+c2e’2v = c1(e") +c2 (ev) =c1x"+c2x”. Section 3.2 14. Imposition of the initial conditions y(0) = O, y'(0) = 0, y"(0) = 3 on the general solution
y(x) = cle’ + czez" + c3e3" yields the three equations c1+c2+c3=1, cl+202+3c3=2, c1+4c2+9c3=0 with solution cl = 3/ 2, c2 = — 3, c3 = 3/ 2. Hence the desired particular solution is
given by y(x) = (3ex — 6e” + 3e3")/2. 21. Imposition of the initial conditions y(0) = 2, y’(0) = — 2 on the general solution
y(x) = c] cosx + c2 sinx + 3x yields the twO equations cl = 2, c2 + 3 = — 2 with
solution c1 = 2, c2 = — 5. Hence the desired particular solution is given by
y(x) = 2 cosx  5 sinx + 3x. 28. If you differentiate the equation c0 + clx + czx2 +~+ cnx" = 0 repeatedly, n times in
succession, the result is the system 2 n _
co+c1x+c2x ++cnx — 0 cl+2c2x++nc,,x"'l = 0 (n—l)!c,,_1+n!cnx = 0 n!c,l = 0 of n+1 equations in the n+1 coefﬁcients co, cl, c2, , c". Upon substitution of x = O, the (k+l)st of these equations reduces to k!ck = 0, so it follows that all these
coefﬁcients must vanish. 37. When we substitute y = vx3 in the given differential equation and simplify, we get the
separable equation xv" + v' = 0 that we solve by writing ——1— => lnv’=—lnx+lnA,
x v"
7 v' = fl— : v(x) = Alnx+B.
x With A =1 and B = O we get v(x) = lnx and hence y2(x) = x3 In x. 38. When we substitute y = vx3 in the given differential equation and simplify, we get the
separable equation xv” + 7v' = O that we solve by writing —v—, = —1 => lnv’ = ~71nx+1nA,
v x
A A
v, = —. 2 = —“"“+B.
x7 VCX) 6x6 With A=—6 and B=0 we get v(x)=l/x6 andhence y2(x)=1/x3. ...
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This homework help was uploaded on 01/21/2008 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell.
 Spring '07
 TERRELL,R
 Math

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