#2 Please, I included #1 for reference. 1. Consider the second-order, constant-coefficient, linear, homogeneous ODEay" + by' + cy = 0where a, b, c ER and a + 0.(a) Solve this equation using the standard method (without a matrix).(b) Find a corresponding system of differential equations and solve again using a matrixequation. For now, assume that the roots are distinct.(c) Show that both solutions are the same.2. Solve ay" + by' + cy =solution in 1(a).0 again using a power series. Show that your solution matches the

q-2 solve ay''+by'+cy=0 using power series let the solution is of the form y=∑n=0∞anxn differentiate with respect to x y'=∑n=1∞nanxn-1 again differentiate with respect to x y''=∑n=2∞nn-1anxn-2substitute the values into the equation a∑n=2∞nn-1anxn-2+b∑n=1∞nanxn-1+c∑n=0∞anxn=0 replace n by n-1 in second term and n by n-2 in first term a∑n=2∞nn-1anxn-2+b∑n=2∞n-1an-1xn-2+c∑n=2∞an-2xn-2=0⇒∑n=2∞ann-1an+bn-1an-1+can-2xn-2=0 hence we get an=-bn-1an-1-can-2a nn-1 for n≥2for n=2 a2=-ba1-ca02·1·a for n=3 a3=-2ba2-ca13·2·a substitute the value of a2 a3=-2b-ba1-ca02·1·a-ca13·2·a =-2b-ba1-ca0-ca112a2 solution is of the form y=a0+a1x+a2x2+a3x3+..... hence we get y=a0+a1x-ba1+ca02·1·ax2-2b-ba1-ca0+ca112a2x3+.....1(a) given equation ay''+by'+cy=0 solve using standard method write the auxillary equation am2+bm+c=0 using quadratic formula , find the roots of the equation m=-b±b2-4ac2a here a≠0 case-1 if b2-4ac=0 we get m=-b2a repeated roots hence solution is of the form y=c1+c2te-b2atif b2-4ac<0 then we get complex roots of the form m=-b2a±pi then solution is of the form y=e-b2atc1cospt+c2sinpt case-iii if b2-4ac>0 then we get m=-b2a±b2-4ac2a then solution is of the form y=c1e-b2a+b2-4ac2at+c2e--b2a-b2-4ac2at