Question

X 8.4.14 Question Help Find the first four nonzero terms in a power series expansion of the solution to the given initial value problem 6y' - 4 e 2xy = 0; y(0)=5 y(x) = (Type an expression that includes all terms up to order 3.)

Answer 1

Griven, Gy' - 4e2x y 0, () YLO) = 5 This is a homogeneous linear differential equation of first order. This variable Separable method, be solved using can from o, 6 dy dae 402% y = 0 22 -> ody da = tény 2x 4 é a da 6 on both sides Integrating dae de y

=) log (y) + log(c) log(c) = 2 et where, log (c) is constant of integration. log (Cy) = 1 e 22 3 2) cy= ereza 3) 2 Yzetes Applying the initial condition, yO=5 Thus, Y(0) = ejeo -) 5- с » c = ek 5 Putting the value of c in X ya cells) ettera 3) y5 y = 1 etera ez xe

2x Žel Y = 5:e Y = 5.e$(828-1) $ Suppose, I 2 (129_1)= u, then y=seu Power series of el es given by, eu - 263 + + It ut uz 21 31 Therefore, power series of y es, y=5eu 263 + » Y=5{1+ 2 + 2? . 31 y = 5+ sut Su? + 5u3 + =) y=5+54 + 5 ži { }(222-13? s{title=4-1} + 628_7%9+

:, y = 5+ 5 (622-1) + 5 z (1-x22) 3 322! + 5 33. 31 (2x _,3 + This is the required power series expansion ycao,