Question

Find the general solution to the following homogeneous differential equations (you answers must have two arbitrary constants (you may use any letters, for example p and q or m and n instead of ki and k2 - notation doesn't matter here]). y" + 2y' – 3y = 0 (b) 6y" – y-y=0 (c) y" + 5y = 0 (d) y" - 9y' +9y 0 The two constants, e.g. ki and k2, determine (and are determined by) an initial condition. When dealing with a second order differential equation, an initial condition is specified by two things: a value of both the function y and its derivative y' are some specific moment, i.e. the data of an initial condition is a pair of values y(ao) = y. and y(20) = S, where to is some specific moment, yo is the value of y at that moment, and s, is the value of the derivative of y at that same moment. Given this additional information, you are able to determine the unique values of the two constants that make everything work out. a 2. Solve the the following differential equations (you answers will not have any arbitrary constants). y" + y - 2y = 0 with y(0) = 1 and y'(0) = 1 (b) 67" - 5y + y = 0 with y(0) 4 and y(0) = 0 (c) y' + 5y + 3y =0 with y(0) = 1 and 7(0) = 0 y" + 8y - 9y=0 with y(1) = 1 and y (1) = 0

2nd order linear homogeneous

Answer 1

SOLUTION a y" +2Y' –3420 D²y + 2 by - 3450 Dad da (0²+2D-3) y =0 Das auxiliary equation m² fam- 320 (m+3)(n-1)=0 mel, -3 Roots are Real and distimet Hence Solution y Cx) = pele + q 234 6 6y"- '-y=0 6D²y-Dy-yzo (60²-0 -D-D geo Auxiliay equation 6m² m-10 6m²_mtam-1 co. 3m ( 4-1) + (2m-1) = 0 (3 m+ 1) (2m-1) zo or & [ Real & distmat Hence y(x) = petx + qe te

=) 8y' + sy' = 0 g²y + Spy=0 (0²+5) y zo Auxiliary Equation m2 +5 to m=&is Rools are complex. Hence solution y (x) = p cosusx + q sinusa d y" – qy' + 99 = 0 2 (0²_qD+q) y co 2ts auxiliary equation m²-qmtq=0 (m-3)2 = m=3, 3 Rools are Real & Repeated Hence y (x) = (p+qx) 3x

SOLUTION 2 y"+y'-2y = 0 D²y + Dy-dyzo (o?+D-2)y co PAS auxiliary equation m2 + m -2=0 metamam-2=0 (m +2) Com-1)=0 Roots ma~2, 1 (Real 4 distinct) Hence g (x) = p ezze +qer but given y (o)=1 1= Ptq Also y'(x) = -2pezze given y'(o) 1 = -2p ta of qer from 0 4 to = 1 Hence y (x) = pe is the solution

1 6y"-sy'ty=0 Pts auxiliay eqh 6m² -smat1=0 6m2-3 mm tico Bm 2m-1)-(am-1)=0 Cam-1) (3m-1) to >> m = 4 4 13 Real 4 disthet Hence y Cx) = pefe + q eft given y(o)=4 4 pta o 4 y' (x) = f et ta ein given y'(O) = 0 Р 0 = + 2 alm 2 from 4 8P + 29=0 »32+2(4-0) co - p=-8 » q =12 Hence g(x) = -8een +12 0

y" + sy' + 3y = 0 auxiliay equation (m2 + sm +3) = 0 m= -52 √25-12 co -SIVB Roots are Real 4 distimet y()= pel-5473) Hence (-5-113) ze tq e 2 given y(o)=1 » 1= pta -0. Also y'(x) = {-s+vi) pe (-S+V13/n pe ut (-5-tis) a key (-Surau given 9(0)30 » Oc (-S+U13) P + (-5-U13) a 03-5 + Ya CP-2) Coo P+Q=1) 1 Allg all V3 CP-9) 5 =p-q (2 Solving 0 P= 5+U13 q= Vis-s 2 V13 Hence 9(x) = (1+345 Je-stu bre + VTB-S le (-s-vile) 2013

d y" +8Y' – 9yco Auxiliay equation m2 +8m - 920 (m +9) (m-1) = 0 M-941 Real t distinct Hence y (X) = p -9% tq ek given Cyc)) = 1 1 = pē9 +qe î p + q é o=e9 O y! (@= -9perkt T trek gjuen g'(0) = 0 o = -9p+q =) 2 -- P=2 9 from 0 4 a a tq e =e9 =) qe9 q= 1+qelo p = e 9 1 tapo Hence y (X) = e94 T+960 99 er Italo

Solved all parts

Any query comment me

Thumbs up!