HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
15.31 Use continuity and the step condition on∂G/∂tatt=t 0 to show that
G(t, t 0 )=α−^1 { 1 −exp[α(t 0 −t)]}for 0≤t 0 ≤t;
x(t)=A(α−a)−^1 {a−^1 [1−exp(−at)]−α−^1 [1−exp(−αt)]}.
15.33 The LHS of the equation is exact for two stages of integration and then needs
an integrating factor expx;2yd^2 y/dx^2 +2ydy/dx+2(dy/dx)^2 ;2ydy/dx+y^2 =
d(y^2 )/dx+y^2 ;y^2 =Aexp(−x)+Bx+C−(sinx−cosx)/2.
15.35 Follow the method of subsection 15.2.6;u(x)=e−x^2 andv(x)satisfiesv′′+4v=
sin 2x, for which a particular integral is (−xcos 2x)/4. The general solution is
y(x)=[Asin 2x+(B−^14 x)cos2x]e−x^2.
15.37 The equation is isobaric, withyof weightm,wherem+p−2=mn;v(x)
satisfiesx^2 v′′+xv′=vn.Setx=etandv(x)=u(t), leading tou′′=unwith
u(0) = 0,u′(0) =λ. Multiply both sides byu′to make the equation exact.