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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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and F(r) = (r - r1)(r - r2). As for second order linear equations with constant coefficients, it is necessary to consider separately the cases in which the roots are real and different, real but equal, and complex conjugates. Indeed, the entire discussion in this section is similar to the treatment of second order linear equations with constant coefficients in Chapter 3 with erx replaced by xr; see also Problem 23.
Real, Distinct Roots. If F(r) = 0 has real roots rx and r2, with rx = r2, then y1 (x) = xr1 and y2(x) = xr2 are solutions of Eq. (1). Since W(xr1, xr2) = (r2 - r1)xr1+r2-1 is nonvanishing for rx = r2 and x > 0, it follows that the general solution of Eq. (1) is
y = c1 Xr1 + c2Xr2, x > 0. (6)
Note that if r is not a rational number, then xr is defined by xr = er ln X.
Solve
2x2y" + 3xy' - y = 0, x > 0. (7)
Substituting y = xr in Eq. (7) gives
xr[2r(r - 1) + 3r - 1] = xr(2r2 + r - 1) = xr(2r - 1)(r + 1) = 0.
Hence rx = 2 and r2 = -1, so the general solution of Eq. (7) is
y = c1 x1/2 + c2x~, x > 0. (8)
Equal Roots. If the roots rx and r2 are equal, then we obtain only one solution y^x) = xr1 of the assumed form. A second solution can be obtained by the method of reduction of order, but for the purpose of our future discussion we consider an alternative method. Since rx = r2, it follows that F(r) = (r - rl)2. Thus in this case not only does F(rx) = 0 but also F'(rx) = 0. This suggests differentiating Eq. (3) with respect to r and then setting r equal to rv Differentiating Eq. (3) with respect to r gives
d L [xr ] = [xrF (r)]. r r
Substituting for F(r), interchanging differentiation with respect to x and with respect to r, and noting that d(xr )/dr = xr ln x , we obtain
L [xr ln x] = (r - r)xr ln x + 2(r - r1)xr. (9)
The right side of Eq. (9) is zero for r = rx; consequently,
y2(x) = xr1 ln x, x > 0 (10)
262
Chapter 5. Series Solutions of Second Order Linear Equations
EXAMPLE
2
is a second solution of Eq. (1). It is easy to show that W(xr1, xr1 ln x) x2r1 \ Hence xr1 and xr1 ln x are linearly independent for x > 0, and the general solution ofEq. (1) is
y (Cj + c2 ln x)xr1, x > 0. (11)
Solve
x2y" + 5xy; + 4y 0, x > 0. (12)
Substituting y xr in Eq. (12) gives
xr [r (r 1) + 5r + 4] xr (r2 + 4r + 4) 0.
Hence r1 r2 2, and
y x-2(Cj + C2lnx), x > 0. (13)
Complex Roots. Finally, suppose that the roots r1 and r2 are complex conjugates, say, r1 X + il and r2 X il, with l 0. We must now explain what is meant by xr when r is complex. Remembering that
xr er ln x (14)
when x > 0 and r is real, we can use this equation to define xr when r is complex. Then
xX+il e(X+iL) lnx eX lnxeiL lnx vXeiL lnx
x e e e x e
xX[cos(x lnx) + i sin(x lnx)], x > 0. (15)
With this definition of xr for complex values of r, it can be verified that the usual laws
of algebra and the differential calculus hold, and hence xr1 and xr2 are indeed solutions ofEq. (1). The general solution ofEq. (1) is
y C1 xX+! L + C2xX-IL. (16)
The disadvantage of this expression is that the functions xX+'L and xX 1L are complex-
2
Xi l
valued. Recall that we had a similar situation for the second order differential equation with constant coefficients when the roots of the characteristic equation were complex. In the same way as we did then we observe that the real and imaginary parts of xX+1L, namely,
xX cos(x lnx) and xX sin(x lnx), (17)
are also solutions ofEq. (1). A straightforward calculation shows that
W[xX cos(xlnx), xX sin(xlnx)] lx2X1.
Hence these solutions are also linearly independent for x > 0, and the general solution ofEq. (1) is
y C1 xX cos(L lnx) + C2xX sin(L lnx), x > 0. (18)
5.5 Euler Equations
263
Solve
3 X + xy' + y = 0. (19)
Substituting y = xr in Eq. (19) gives
xr [r (r 1) + r + 1] = xr (r2 + 1) = 0.
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