Find the first four nonzero terms in a power series expansion of the solution to the given initial value problem. 6y'-5e²xy=0; y(0) = 1 ∞ Rewrite the given differential equation in terms of a power series, using a general solution y(x)= Σ an (x-xo)". Expand e²x in a power series, writing the first four terms. Select the correct n=0 answer below and fill in the answer box to complete your choice. Ο.Α. 6 Σ n(n-1)an (x-xο) "-2-5( n=2 +...) Σ an(x-x)" =0 n=0 00 00 ○ B. 6 Σ n(n-1)a (x-x)n-²-5( _ +...) Σ na (x-xò) ª−1=0 n=2 n=1 O c. 6(+) Σ nan (x-xo)n-1-5 Σan (x-x)=0 n=1 n=0 ∞ ✅D. 6 ¶ na, (x-x)n−1−5[1+2x+ 2x² + 3x³ +..) [ ³₂ (x-x)² =0 Σ Σan n=1 n=0 y(x)=+ (Type an expression that includes all terms up to order 3.) + ...

Consider the differential equation x^2y" + x (1−x)y' − xy = 0. a. Show that X0=0 is a regular singular point. b. Determine the indicial equation, the recurrence relation, and the roots of the indicial equation. c. Find the series solution for x>0 corresponding to the larger root. d. Find the series solution corresponding to the smaller root by following the procedure outlined in section 5.4 and demonstrated in section 5.7.

Find a power series expansion about x = 0 for a general solution to the given differential equation. Your answer should include a general formula for the coefficients. -z" -x²z' - xz = 0 What is the power series solution to the differential equation? A. Z(x) = a 1 + B. Z(X) = a1 + ∞ Σ (-1) - n=1 (1.4.7... (3n - 2))² (3n)! ∞ Σ (-1). n=1 ∞ OC. Z(x) = a 1+ Σ n=1 D. Z(X) =a 1+ Σ n=1 (3.6.9...(3n)) (3n - 2)! (1.4.7... (3n - 2))² (3n)! -x³n (3.6.9...(3n)) (3n - 2)! 3n 3n + a X + 2,³^) - ₁,[(x + 2, + x+ Σ n=1 X + x+ Σ n=1 ∞ n=1 (-1)^. ∞ Σ (-1). n=1 (2.5.8...(3n-1))² (3n + 1)! (1.4.7...(3n-2)) +1 (3n + 2)! -x³n+ (2.5.8--- (3n-1))² (3n+ 1)! (1.4.7...(3n-2)) (3n+2)! -x³n+1 -x³n+1 -x³n+ 1

Diff eqns