Answers to Odd-Numbered Exercises 445
(iv) This is true becauseu 1 (x)andu 2 (x)are solutions of the homoge-
neous equation.Chapter 0 Miscellaneous Exercises
1.u(x)=T 0 cosh(γx)+(
T 1 −T 0 cosh(γa))sinh(γx)
sinh(γa).
3.u(x)=T 0.
5.u(r)=p(a^2 −r^2 )/4.
7.u(ρ)=H(a^2 −ρ^2 )/ 6 +T 0.
9.u(x)=T+(T 1 −T)cosh(γx)/cosh(γa).
11.u(x)=T 0 +(T−T 0 )e−γx.
13.h(x)=√
ex(a−x)+h^20 +(h^21 −h^20 )(x/a).
15.u(x)=w( 1 −e−γxcos(γx))EI/k,whereγ=(k/ 4 EI)^1 /^4.17.u(x)={
T 0 +Ax, 0 <x<αa,
T 1 −B(a−x), αa<x<a,A=κ 2
κ 1 ( 1 −α)+κ 2 αT 1 −T 0
a , B=κ 1
κ 2 A.19.u(x)=^1
2(
1 −e−^2 x)
−^1
2
(
1 −e−^2 a) 1 −e−x
1 −e−a.
- a.u(x)=sinh(px)/sinh(pa);
b.u(x)=cosh(px)−cosh(pa)
sinh(pa)sinh(px)=sinh(
p(a−x))
/sinh(pa);c.u(x)=cosh(px)/cosh(pa);
d.u(x)=cosh(p(a−x))/cosh(pa);
e.u(x)=−cosh(p(a−x))/psinh(pa);
f.u(x)=cosh(px)/psinh(pa).23.u(x)=x
2 ln∣∣
∣∣^1 +x
1 −x∣∣
∣∣−1.
- Multiply byu′and integrate:^12 (u′)^2 =^15 γ^2 u^5 +c 1 .Sinceu(x)→ 0
asx→∞,alsou′(x)→0; thusc 1 =0. Nowu′=−
√
2 γ^2 / 5 u^5 /^2 or
u−^5 /^2 u′=−√
2 γ^2 /5(thenegativerootmakesudecrease) can be in-
tegrated to result in(− 2 / 3 )u−^3 /^2 =−√
2 γ^2 / 5 x+c 2. The condition at
x=0givesc 2 =(− 3 / 2 )U−^3 /^2.