2) Suppose you have a differential equation that satisfies the conditions of Theorem 6.2.1- Existence of Power Series Solutions with center x, = 0. It is then guaranteed that a solution exists of the form y(x) = E-0 Cnx". The recurrence relation for the coefficients was calculated to be k 3 Ck+2 k = 1,2,3, .. 2(k + 2) k+1 2(k + 2)(k + 1) 3 C2 = -7C0 Find the solution y(x) to the initial-value problem if y(0) = 0 and y'(0) = 1. Give at least the first 3 nonzero terms of the solution.

Use the procedure in Example 8 in Section 6.2 to find two power series solutions of the given differential equation about the ordinary point x = 0. y' + e'y' – y = 0 1,3 ܐ ܐܨܐ ܀ Oy=1+ Oy, =1+ 2 Oy =1+ 2 ܠܐܕܐܐܨܐ+Oy + + .… and y2 - X - and y, = x + Oy:=1+_x+x + 2 3 and y, = x - +ܐ 4 ܡܐ ܠܨ-xܕx- + y+1-:0 ܕܨܐ y2 X + 13 and y2 = x 6 + ܟܛܚ ܕ - -x- ܕOy -1- o- 3 - .… and y ܐܐܐܐ 4 9 1 24 - 9 1 24 ܐ ܐ ܐ x- + .… 1 16 1x2 + 1x³ + 1 16

In solving a differential equation by Frobenius method of series solution about x = 0, we obtain the following indicial equation and recurrence relation: 7(2r–1)=0 and 1 Cn+1 = %3D 6n +r+3n where n = 0,1,2,3,... Find the solution corresponding to the larger root of the indicial equation. Include the first three nonzero terms, use Co=I O A. Y = 1+x 4 x+... 133 O B. Y =x-1/2/ 4 3 +.. 133 O c.y =xl/2| 12 4 3 +... c. 4 2 O D. y = x1/2 1+x+ 133 ++.. O E.Y =x1/2 1+x+ 19 4 2 +...

2- Consider the differential equation (²+3)y" - 4y' = 0. The recurrence relation for the coefficients an of the power series solution an" about the ordinary point o=0 is given by b) an+2= an+2= an+2= none of these an+2= 12-0 4nan+1 + (n² = n)an 3n² +9n+6 n22 (4n+4)an+1-(n² - n)an 3n² +9n+6 (n + 4)an+1 +n²an 3n² +9n+6 an+1 + (n²-n)an 3n² +9n+6 1 " n>2 n≥ 2 n22