Q29E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to  $y^{\left(4\right)}+2y^{\left(3\right)}+4y\text{'}\text{'}+3y\text{'}+2y=0$ by using Newton’s method to approximate numerically the roots of the auxiliary equation. [Hint: To find complex roots, use the Newton recursion formula $z_{n+1}=z_n-\frac{f\left(z_n\right)}{f\text{'}\left(z_n\right)}$ and start with a complex initial guess z0.]

Step-by-Step Solution

Answer

_{The general solution is $y\left(x\right)=c_1{e}^{-0,5x}\cos (0,86602x)+c_2{e}^{-0,5x}\sin (0,86602x)+c_3{e}^{-0,5x}\cos (1,32288x)+c_4{e}^{-0,5x}\sin (1,32288x)$}

Step 1: Newton’s Approximation method

Newton's Method, also known as Newton Method, is important because it's an iterative process that can approximate solutions to an equation with incredible accuracy. And it's a method to approximate numerical solutions (i.e., x-intercepts, zeros, or roots) to equations that are too hard for us to solve by hand.

Step 2: Use of Newton’s Approximation method

We are going to find the roots of auxiliary equation by using Newton’s Approximation method :

$\begin{array}{l}{r}^4+2r^3+4r^2+3r+2=0\\ f\left(x\right)=x^4+2x^3+4x^2+3x+2\\ f\text{'}\left(x\right)=4x^3+6x^2+8x+3\\ \\ z_{n+1}=z_n-\frac{f\left(z_n\right)}{f\text{'}\left(z_n\right)},n=1,2,3,\mathrm{...}\\ z_{n+1}=z_n-\frac{{z_n}^4+2z_n^3+4z_n^2+3z_n+2}{4z_n^3+6z_n^2+8z_n+3},n=1,2,3,\mathrm{...}\\ \\ z_2=0.015+0,48i\\ z_3=-0,41023+0,60467i\\ z_4=-0,50475+0,84548i\\ z_5=-0,50000+0,86561i\\ z_6=-0,50000+0,86602i\\ z_7=-0,5+0,86602i\\ \\ z_{n+1}=z_n-\frac{{z_n}^4+2z_n^3+4z_n^2+3z_n+2}{4z_n^3+6z_n^2+8z_n+3},n=1,2,3,\mathrm{...}\\ z_1=-0,5+1,5i\\ z_2=-0,5+1,375i\\ z_3=-0,5+1,329481i\\ z_4=-0,49999+1,323i\\ z_5=-0,5+1,32288i\\ z_6=-0,5+1,32288i\\ \\ y\left(x\right)=c_1{e}^{-0,5x}\cos (0,86602x)+c_2{e}^{-0,5x}\sin (0,86602x)+c_3{e}^{-0,5x}\cos (1,32288x)+c_4{e}^{-0,5x}\sin (1,32288x)\end{array}$

Hence, the final answer is :

 

_{$y\left(x\right)=c_1{e}^{-0,5x}\cos (0,86602x)+c_2{e}^{-0,5x}\sin (0,86602x)+c_3{e}^{-0,5x}\cos (1,32288x)+c_4{e}^{-0,5x}\sin (1,32288x)$}