Question

Now you start 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 + Binz = 0, subject to x(0) = 0 and x'0) = ; 20. Discuss conditions under which your reasoning is accurate, and make sure to include everything you can say. Make choices for 2 to illustrate your points ONLY; make sure to address the problem in full generality in a separate conclusive paragraph. dla

Answer 1

Solution:-

Given that

The Taylor series of a function is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point.

Given

dr + sinc dt2 0

subject to
(0) = 0
and
I'(0) = 11 > 0

Now we are considering fact that Taylor series can be used to steady approximate solutions to non linear ODE's

Let us take

α(t) =Σα, " =0

then we get

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

similarly

$x''(t)=\sum^\infty_{n=2}n(n-1)a_nt^{n-2}$

By applying

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$

Thus we get

$\sum^\infty_{n=2}n(n-1)a_nt^{n-2}+\sum^\infty_{n=0}a_nt^n-\frac{(\sum^\infty_{n=0}a_nt^n)^3}{3!}+\frac{(\sum^\infty_{n=0}a_nt^n)^5}{5!}=0$we already know that from equation

(0) = 0

then we take

Ор = 0

Now substitute in the above equation

Ор = 0

$\sum^\infty_{n=2}n(n-1)a_nt^{n-2}+\sum^\infty_{n=0}a_nt^n-\frac{t^3(\sum^\infty_{n=0}a_nt^{n-1})^3}{6}+\frac{t^5(\sum^\infty_{n=0}a_nt^{n-1})^5}{120}...=0$Now considering constant terms as

2(1)a2 = 0

$x'(0)=x_1=a_1$

Now by considering coefficient of t

we have

$3(2)a_3+a_1=0$

By solving

$6a_3+a_1=0$

$6a_3=-a_1$

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

we already know that $x_1=a_1$ substitute

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

Now considering coefficient of $t^2$

$12a_4+a_2=0$

we get

Now considering coefficient of $t^3$

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

then we get

$20a_5-\frac{2a_1}{6}=0$

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

as we know that $a_1=x_1$ then finally we get

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

Now similarly considering coefficient of $t^4$

$30a_6+a_4-\frac{3a^2_1a_2}{6}=0$

we already know that $a_2=0$ and $a_4=0$

Thus we get

$a_6=0$

Therefore

$x(t)=x_1t-\frac{x_1}{6}t^3+\frac{x_1}{60}t^5...$

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