# 9780470458365, Chapter 3.1, Problem 2P

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Chapter 3.1, Problem 2P Step-by-step solution Step1072 Consider the functions ¢* ¢ * ¢** and the ODE y™—2)"—p'+2y=0 ... 1) The objective is to show that the functions ¢*,¢*, ¢ are solutions and form a basis on any interval by using Wronskians. First, show that the given functions are solutions of equation (1). In the equation (1), the ODE is a third order homogeneous linear ODE with constant coefficients. Therefore, the corresponding characteristic equation of (1), is given as follows: A=227-242=0 By inspection 2 =1 is a root of the characteristic equation. Since |* 2(1)' ~1+2=0- Hence, (A1) is a factor, so divide ;* 3% 7 +2 by 2] tofind the other roots. A-2-2 AP 20222 -(2'+2%) -A=2 -(2*+2) -22+2 -(-24+2) 0 Thus, 2'-22"-2+2=(A-1)(A*-2-2)=0 Theroots of ;2 _,_2 - are A=-1,2. By inspection. Therefore, the three roots of 33 2,2 _j4+2=0 are A=1,~land2. The roots are real and distinct hence, the solution of (1) is l y=ce' +ee +c e Therefore, the functions ¢* ¢ ¢** are solutions of ODE (1). s Step 2072 The ODE has continuous coefficients, so apply the following stated theorem: Theorem : Letthe ODE y'" + p, (x)y"" " +..+ p,(x) '+ p, (x) y =0 ... (2) have the coefficients p (x) .... p,, (x) on an open interval I. Then » solutions of (2) on | are linearly dependent on | if and only if their Wronskian is zero for some x = x, inl. Furthermore, if there is an x, in | at which Wronskian is not zero, then y, ..... ,y, are linearly independent on I, so that they form a basis of solutions of (2) on I. Now, to apply this theorem, compute the Wronskian of the functions ¢*,¢™*,¢** as follows: Let y, =¢',y,=¢"and y, =¢™ Thus, yl' =e", y" =e", y:' =—e ", yz' =e ", y; =2¢** and y; = 4™ N V2 Vs W (32 05) =91 25 2 wov:on Substituting the values to get the following: o et o x x 2\ _| x_ _-x 52 Wle',e e )=le"—e " 2e o et 4et -x 2x I e' e =e*[l —e 2¢| Factorout . from first column. 1 e 4¢™ 1 1é™ =e'e*|l -1 2¢*| Factorout .+ from second column. 11 4e™ I 11 =e |l -1 2| Factor out 2+ from third column. 1 14 =e* [l(—4 -2)-1(4-2)+1(1+ l)] Expand the determinant. =—be" Exponential function never zero. #0 forall x Hence, by above stated theorem, the functions ¢* ¢ *,¢** are linearly independent and so they form a basis of solutions of the ODE y"—-2y"—y'+2y=0.
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