Question

It's a fact that Taylor series can be used to study approximate solutions to non-linear ODEs. Use this fact to study approximate solutions to dec + sin x = 0, dt2 (1) subject to x(0) = 0 and x'(0) = x1 > 0. Discuss conditions under which your reasoning is accurate, and make sure to include everything you can say. Make choices for zı to illustrate your points ONLY; make sure to address the problem in full generality in a separate conclusive paragraph.

Answer 1

Solution:-

Given that

dr + sinc dt2 0

here,

Now here,

Let

α(t) =Σα, " =0

$x'(t)=\sum^\infty_{n=1}na_n t^{n-1}$

2"(t) nan 1)antn-2 n=2

using

sin I = 1 3! -+ 5!

$\sum^\infty_{n=2}n(n-1)a_n t^{n-2}+\sum^\infty_{n=0}a_nt^n-\frac{(\sum^\infty_{n=0}a_nt^n)^3}{3!}+\frac{(\sum^\infty_{n=0}a_nt^n)^5}{5!}..=0$Now here,

Given

(0) = 0

So, ,

Ор = 0

So we get

$\sum^\infty_{n=2}n(n-1)a_n t^{n-2}+t\sum^\infty_{n=1}a_nt^{n-1}-\frac{t^3(\sum^\infty_{n=1}a_nt^{n-1})^3}{6}+\frac{t^5(\sum^\infty_{n=1}a_nt^{n-1})^5}{120}..=0$Now By comparing constant term

2(1)a2 = 0

Tp = lx = (0,2

comparing coefficient of t

3(2) az + a1 = 0

ნll = —l1

$a_3=\frac{-a_1}{6}=\frac{-x_1}{6}$

comparing coefficient of $t^2$

$12a_4+a_2=0$

$a_4=0$

comparing coefficient of $t^3$

$20a_5+a_3-\frac{a_1}{6}=0$

2005 201 6 = 0

$a_5=\frac{a_1}{60}=\frac{x_1}{60}$

Now, comparing coefficient of $t^4$

$30a_6+a_4-\frac{3a_1^2a_2}{6}=0$

16 = 0
(Since $a_2=a_4=0$ )

$x(t)=x_1-\frac{x_1}{6}t^3+\frac{x_1}{60}t^5...$

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