Answer:
[tex]\mathbf{y(t) = t + \dfrac{7}{2}t^2 - \dfrac{2}{3}t^3+ ...}[/tex]
Step-by-step explanation:
THe interpretation of the given question is as follows:
y'' + ky + ry³ = A cos ωt
Let k = 4, r = 3, A = 7 and ω = 8
The objective is to find the first three non zero terms in the Taylor polynomial approximation to the solution with initial values y(0) = 0 ; y' (0) = 1
SO;
y'' + ky " ry³ = A cos ωt
where;
k = 4, r = 3, A = 7 and ω = 8
y(0) = 0 ; y' (0) = 1
y'' + 4y + 3y³ = 7 cos 8t
y'' = - 4y - 3y³ + 7 cos 8t ---- (1)
∴
y'' (0) = -4y(0) - 3y³(0) + 7 cos (0)
y'' (0) = - 4 × 0 - 3 × 0 + 7
y'' (0) = 7
Differentiating equation (1) with respect to t ; we have:
y''' = - 4y' - 9y² × y¹ - 56 sin 8t
y''' (0) = -4y'(0) - 9y²(0)× y¹ (0) - 56 sin (0)
y''' (0) = - 4 × 1 - 9 × 0 × 1 - 56 × 0
y''' (0) = - 4
Thus; we have :
y(0) = 0 ; y'(0) = 1 ; y'' (0) = 7 ; y'''(0) = -4
Therefore; the Taylor polynomial approximation to the first three nonzero terms is :
[tex]y(t) = y(0) + y'(0) t + y''(0) \dfrac{t^2}{2!} + y'''(0) \dfrac{t^3}{3!}+...[/tex]
[tex]y(t) = 0 + t + 7 \dfrac{t^2}{2!} + \dfrac{-4}{3!} {t^3}+ ...[/tex]
[tex]\mathbf{y(t) = t + \dfrac{7}{2}t^2 - \dfrac{2}{3}t^3+ ...}[/tex]
