Variable coefficients second order linear ODE (Sect. 2.1 ...

1 Variable coefficients second order linear ODE (Sect. ).ISecond order linear and uniqueness of dependent and independent Wronskian of two and fundamental s theorem on the second order nonlinear order linear differential functionsa1,a0,b:R R, the differential equation in theunknown functiony:R Rgiven byy +a1(t)y +a0(t)y=b(t)(1)is called asecond order lineardifferential equation equation in (1) is calledhomogeneousiff for allt Rholdsb(t) = equation in (1) is called ofconstant coefficientsiffa1,a0, andbare :The notion of an homogeneous equation presented hereis not the same as the notion presented in the previous order linear differential (a)A second order , linear , homogeneous, constant coefficientsequation isy + 5y + 6 = 0.

2 (b)A second order order , linear , constant coefficients,non-homogeneous equation isy 3y +y= 1.(c)A second order , linear , non-homogeneous, Variable coefficientsequation isy + 2t y ln(t)y=e3t.(d)Newton s second law of motion (ma=f) for point particles ofmassmmoving in one space dimension under a forcef:R Ris given bym y (t) =f(t).CVariable coefficients second order linear ODE (Sect. ).ISecond order linear and uniqueness of dependent and independent Wronskian of two and fundamental s theorem on the second order nonlinear the functions y1and y2are solutions to the homogeneous linearequationy +a1(t)y +a0(t)y= 0,(2)then the linear combination c1y1(t) +c2y2(t)is also a solution forany constants c1, c2 :Verify that the functiony=c1y1+c2y2satisfies Eq.

3 (2) forevery constantsc1,c2,that is,(c1y1+c2y2) +a1(t)(c1y1+c2y2) +a0(t)(c1y1+c2y2)= (c1y 1+c2y 2) +a1(t)(c1y 1+c2y 2) +a0(t)(c1y1+c2y2)=c1[y 1+a1(t)y 1+a0(t)y1]+c2[y 2+a1(t)y 2+a0(t)y2]= coefficients second order linear ODE (Sect. ).ISecond order linear and uniqueness of dependent and independent Wronskian of two and fundamental s theorem on the second order nonlinear and uniqueness of ( Variable coefficients)If the functions a,b: (t1,t2) Rare continuous, the constantst0 (t1,t2)and y0, y1 R, then there exists a unique solutiony: (t1,t2) Rto the initial value problemy +a1(t)y +a0(t)y=b(t),y(t0) =y0,y (t0) =.

4 IUnlike the first order linear ODE where we have an explicitexpression for the solution, there isno explicit expressionforthe solution of second order linear integrationsmust be done to find solutions tosecondorder linear . Therefore, initial value problems withtwo initialconditionscan have a unique and uniqueness of the longest intervalI Rsuch that there exists a uniquesolution to the initial value problem(t 1)y 3ty + 4y=t(t 1),y( 2) = 2,y ( 2) = :We first write the equation above in the form given inthe Theorem above,y 3tt 1y +4t 1y= intervals where the hypotheses in the Theorem above aresatisfied, that is, where the equation coefficients are continuous,areI1= ( ,1) andI2= (1, ).

5 Since the initial conditionbelongs toI1, the solution domain isI1= ( ,1).CExistence and uniqueness of :IEvery solution of the first order linear equationy +a(t)y= 0is given byy(t) =c e A(t),withA(t) = a(t) solutions above are proportional to each other:y1(t) =c1e A(t),y2(t) =c2e A(t) y1(t) =c1c2y2(t)Remark:The above statement isnot truefor solutions of secondorder, linear , homogeneous equations ,y +a1(t)y +a0(t)y= we prove this statement we need few definitions:IProportional functions (linearly dependent).IWronskian of two coefficients second order linear ODE (Sect. ).ISecond order linear and uniqueness of dependent and independent Wronskian of two and fundamental s theorem on the second order nonlinear dependent and independent continuous functionsy1,y2: (t1,t2) R Rare calledlinearly dependent, (ld),on the interval (t1,t2) iff there exists aconstantcsuch that for allt Iholdsy1(t) =c y2(t).

6 The two functions are calledlinearly independent, (li),on theinterval (t1,t2) iff they are not linearly :Iy1,y2: (t1,t2) Rare ld there exist constantsc1,c2, notboth zero, such thatc1y1(t) +c2y2(t) = 0for allt (t1,t2).Iy1,y2: (t1,t2) Rare li the only constantsc1,c2, solutionsofc1y1(t) +c2y2(t) = 0for allt (t1,t2) arec1=c2= dependent and independent (a)Show thaty1(t) = sin(t),y2(t) = 2 sin(t) are ld.(b)Show thaty1(t) = sin(t),y2(t) =tsin(t) are :Case (a) (b):Find constantsc1,c2such that for allt Rholdsc1sin(t) +c2tsin(t) = 0 (c1+c2t) sin(t) = att= /2 andt= 3 /2 we obtainc1+ 2c2= 0,c1+3 2c2= 0 c1= 0,c2= conclude:The functionsy1andy2are coefficients second order linear ODE (Sect.)

7 ISecond order linear and uniqueness of dependent and independent Wronskian of two and fundamental s theorem on the second order nonlinear Wronskian of two :The Wronskian is a function that determines whethertwo functions are ld or functionsy1,y2: (t1,t2) Ris the functionWy1y2(t) =y1(t)y 2(t) y 1(t)y2(t).Remark:IIfA(t) =[y1y2y 1y 2],thenWy1y2(t) = det(A(t)).IAn alternative notation is:Wy1y2= y1y2y 1y 2 .The Wronskian of two the Wronskian of the functions:(a)y1(t) = sin(t) andy2(t) = 2 sin(t).(ld)(b)y1(t) = sin(t) andy2(t) =tsin(t).(li)Solution:Case (a):Wy1y2= y1y2y 1y 2 = sin(t) 2 sin(t)cos(t) 2 cos(t).

8 Therefore,Wy1y2(t) = sin(t)2 cos(t) cos(t)2 sin(t) Wy1y2(t) = (b):Wy1y2= sin(t)tsin(t)cos(t) sin(t) +tcos(t) .Therefore,Wy1y2(t) = sin(t)[sin(t) +tcos(t)] cos(t)tsin(t).We obtainWy1y2(t) = sin2(t).CThe Wronskian of two :The Wronskian determines whether two functions arelinearly dependent or (Wronskian and linearly dependence)The continuously differentiable functions y1, y2: (t1,t2) Rarelinearly dependent iffWy1y2(t) = 0for all t (t1,t2).Remark:Importance of the Wronskian:ISometimes it is not simple to decide whether two functionsare proportional to each Wronskian is useful to study properties of solutions toODE without having the explicit expressions of thesesolutions.

9 (See Abel s Theorem later on.)The Wronskian of two whether the following two functions form a or set:y1(t) = cos(2t) 2 cos2(t),y2(t) = cos(2t) + 2 sin2(t).Solution:Compute their Wronskian:Wy1y2(t) =y1y 2 y (t) =[cos(2t) 2 cos2(t)] [ 2 sin(2t) + 4 sin(t) cos(t)] [ 2 sin(2t) + 4 sin(t) cos(t)] [cos(2t) + 2 sin2(t)].sin(2t) = 2 sin(t) cos(t) [ 2 sin(2t) + 4 sin(t) cos(t)]= concludeWy1y2(t) = 0, so the coefficients second order linear ODE (Sect. ).ISecond order linear and uniqueness of dependent and independent Wronskian of two and fundamental s theorem on the second order nonlinear and fundamental a1, a0: (t1,t2) Rare continuous, then the functionsy1,y2: (t1,t2) Rsolutions of the initial value problemsy 1+a1(t)y 1+a0(t)y1= 0,y1(0) = 1,y 1(0) = 0,y 2+a1(t)y 2+a0(t)y2= 0,y2(0) = 0,y 2(0) = 1,are linearly.

10 IEvery linear combinationy(t) =c1y1(t) +c2y2(t), is also asolution of the differential equationy +a1(t)y +a0(t)y= 0,IConversely, every solutionyof the equation above can bewritten as a linear combination of the solutionsy1, and fundamental :The results above justify the following solutionsy1,y2of the homogeneous equationy +a1(t)y +a0(t)y= 0,(3)are calledfundamental solutionsiff the functionsy1,y2are linearlyindependent, that is, iffWy1y26= any two fundamental solutionsy1,y2, and arbitrary constantsc1,c2, the functiony(t) =c1y1(t) +c2y2(t)is called thegeneral solutionof Eq. (3).General and fundamental thaty1= tandy2= 1/tare fundamental solutions of2t2y + 3t y y= :First show thaty1is a solution:y1=t1/2,y 1=12t 1/2,y 1= 14t 3/2,2t2( 14t 32)+ 3t(12t 12) t12= 12t12+32t12 t12= show thaty2is a solution:y2=t 1,y 2= t 2,y 2= 2t 3,2t2(2t 3)+ 3t( t 2) t 1= 4t 1 3t 1 t 1= and fundamental thaty1= tandy2= 1/tare fundamental solutions of2t2y + 3t y y= :We show thaty1,y2are linearly (t) = y1y2y 1y 2 = t1/2t 112t 1/2 t 2.