Question

In the study of a nonlinear spring with periodic​ forcing, the equation y prime prime plus ky plus ry cubedy′′+ky+ry3equals=Upper A cosine omega tAcosωt arises. Let kequals=44​, requals=33​, Aequals=77​, and omegaωequals=88. Find the first three nonzero terms in the Taylor polynomial approximation to the solution with initial values ​y(0)equals=​0, y prime (0 )y′(0)equals=1.

Answer 1

Answer:

$\mathbf{y(t) = t + \dfrac{7}{2}t^2 - \dfrac{2}{3}t^3+ ...}$

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 :

$y(t) = y(0) + y'(0) t + y''(0) \dfrac{t^2}{2!} + y'''(0) \dfrac{t^3}{3!}+...$

$y(t) = 0 + t + 7 \dfrac{t^2}{2!} + \dfrac{-4}{3!} {t^3}+ ...$

$\mathbf{y(t) = t + \dfrac{7}{2}t^2 - \dfrac{2}{3}t^3+ ...}$