Solve each initial-value problem. Express each answer in the form of Eq. 1.6.10, 1.6.13, or 1.6.19. 33. u" 9u = 0, u (0) = 0, u'(0) = 1 34. u" 5u' + 6u = 0, u (0) = 2, u'(0) = 0 35. u" +4u' + 4u = 0, u (0) = 0, u'(0) = 2 36. u4u = 0, u(0) = 2, u'(0) = 1 A general solution is u(x) = c₁ex/2 + C₂xe¯ax/2 = (c₁ + C2x)e-ax/2 Note that, using u₁ = eax/2 and u₂ = xeax/2, the Wronskian is (1.6.19) 121 Using the appropriate identities, 12 this solution can be put in the following two equivalent forms: u(x)=eTax/2[A sinh (Va2 – 4bx) + B cosh (V@2 – 4bx)] u(x) = cewx/2 sinh (;Va? – 4bx+ca) (1.6.10) (1.6.11) which, with the appropriate identities, 13 can be put in the following two equivalent forms: u(x)=ex/[A sin (√a² - 4bx) + B cos (½√a² - 4bx)] u(x)=c3e e-ax/2 cos (√√a²-4bx + c4) (1.6.13) (1.6.14) 1.6 HOMOGENEOUS, SECOND-ORDER, LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 39 Let us consider two cases (a²-4b) > 0 and (a²-4b) <0. If (a-4b) > 0, the solution takes the form (1.6.9) Using the appropriate identities, this solution can be put in the following two equivalent forms: (x) [A sinh (√a²-4bx)+ B cosh (√a²-4b.x)] u(x)=cesinh (√a²-4bx+c) If (a²-4b) < 0, the general solution takes the form, using i = √-I. (x)=(cie (1.6.10) (1.6.11) (1.6.12) which, with the appropriate identities, can be put in the following two equivalent forms: (x)=[A sin (√a-4bx)+ B cos (√a²-4bx)] u(x)=cyecos (Va²-4bx+cs) (1.6.13) (1.6.14) If a particular form of the solution is not requested, the form of Eq. 1.6.9 is used if (a² - 4b) > 0 and the form of Eq. 1.6.13 is used if (a-4b) <0. If (a²-4b)=0,mmy and a double root occurs. For this case the solution 1.6.8 no longer is a general solution. What this means is that the assumption that there are two linearly independent solutions of Eq. 1.6.1 of the form is false, an obvious conclusion since W(x)=0. To find a second solution we make the assumption that it is of the form 2(x)=(x) where m² + am + b = 0. Substitute into Eq. 1.6.1 and we have (m²+ame + be)+(2me" + a) + dv dx (1.6.15) =0 dr² (1.6.16) The coefficient of v is zero since m²+am+b=0. The coefficient of du/dx is zero since we are assuming that m² +am+b=0 has equal roots, that is, m = -a/2. Hence, 13The appropriate identities are The appropriate identities are d² =0 dx² eason 1 + sinh x sinh(x + y) = sinh x cosh y + sinh y cosh x cos+isin costar+B)=cosa cos - sina sin cos² + sin² = 1 (1.6.17)