homework 07 solutions

homework 07 solutions - Section 3.1 19 21 30 31 51 If y=1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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 =C|x3| 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. ...
View Full Document

{[ snackBarMessage ]}

Page1 / 2

homework 07 solutions - Section 3.1 19 21 30 31 51 If y=1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online