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.]

Short 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-by-step solution

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.

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 :

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