#2 Please. I can get to the recursive function, but I need help figuring out how to match the recursive function to 1a 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

Given: Let's by using a power series, solve ay" + by' + cy = 0 once more. Demonstrate that your solution is the same as the one in 1 (a).Solution: Let's consider the general solution of the given ay" + by' + cy = 0 is y(x)=Aem1x+Bem2x-----1 Let us substitute y=∑n=0∞enxn, yι=∑n=1∞n1enxn-1, yιι=∑n=2∞nn-1enxn-2 Let's put it all together, a×∑n=2∞nn-1enxn-2+b×∑n=1∞nenxn-1+c×∑n=0∞enxn=0 ⇒a×∑n=0∞n+2n+1en+2xn+b×∑n=0∞n+1en+1xn+c×∑n=0∞enxn=0 Let's take the coefficient of xn=0, ⇒a×n+2n+1en+2+b×n+1en+1+c×en=0Let's put n=0,then ⇒2a×e2+b×e1+c×e0=0 ⇒e2=-be12-ce02 let's put n=1, ⇒a×e3+2b×e2+c×e1=0 ⇒e3=-bc13ae2-ce16a =-b3a-6e12-ce02-ce16a =b2-c6ae1+bc6ae0let's put n=2, ⇒12a×e4+3b×e3+c×e2=0 e4=-b4ae3-c12ac2 =-b4ab2-c6ae1+bc6ae0-e12a-bc12-ce02 =-bb2-c24a2+bc24a2e1+-b2c24a2+c224ae0 Thus the solution must be, yn=e0+e1x+e2x2+e3x3+e4x4+...=e01-c2x2+bc4ax3+b2c24a2+c224ax4+...+e1x-b2x2+b2-c6ax3+-bb2-c24a2+bc24ax4+... =c0×em1x+c1×em2x-----2 let's replace c0=A and c1=B ∴=Aem1x+Bem2x Thus equation 1= equation 2Answer: Therefore the solution of the given equation ay" + by' + cy = 0 by using power series matches the solution using the standard method. y(x)=Aem1x+Bem2x Hence proved.