91Ó°ÊÓ

Given:

The differential equation is$ 3y\text{'}\text{'}\text{'}+18y\text{'}\text{'}+13y\text{'}-19y=0$

The auxiliary equation for the above equation$3m^3+18m^2+13m-19=0$

Let,

$$\begin{gathered} f\left(x\right)=3x^3+18x^2+13x-19 \\ f\text{'}\left(x\right)=9x^2+36x+13 \end{gathered}$$

One knows that, Newton’s formula.

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f\text{'}\left(x_n\right)}$

Now, Using Newton’s method to find the roots of the auxiliary equation,

Assume, ${\mathrm{x}}_0=0.5$.

$$\begin{gathered} f\left(x_0\right)=3{x_0}^3+18{x_0}^2+13x_0-19 \\ f\text{'}\left(x_0\right)=9{x_0}^2+36x_0+13 \end{gathered}$$

Firstly, solve for x.

$$\begin{gathered} x_1=x_0-\frac{f\left(x_0\right)}{f\text{'}\left(x_0\right)} \\ {x}_1=\left(0.5\right)-\frac{3{\left(0.5\right)}^3+18{\left(0.5\right)}^2+13\left(0.5\right)-19\left(0.5\right)}{9{\left(0.5\right)}^2+36\left(0.5\right)+13} \\ {x}_1=\left(0.5\right)-\frac{\left(-7.625\right)}{33.25} \\ {x}_1=0.7291 \end{gathered}$$

Similarly,

$$\begin{gathered} x_1=0.7291682113606 \\ {x}_2=0.701646387314951 \\ {x}_3=0.701196457929202 \\ {x}_4=0.701196086359434 \\ {x}_5=0.701196086147743 \\ {x}_6=0.701196086147623 \\ {x}_7=0.701196086147623 \end{gathered}$$

Values are approximately equal.

The root is $m_1=0.701190$.

Assume, $x_0=-1$.

$$\begin{gathered} f\left(x_0\right)=3{x_0}^3+18{x_0}^2+13x_0-19 \\ f\text{'}\left(x_0\right)=9{x_0}^2+36x_0+13 \end{gathered}$$

Firstly, solve for $x_1$.

$$\begin{gathered} x_1=x_0-\frac{f\left(x_0\right)}{f\text{'}\left(x_0\right)} \\ {x}_1=\left(-1\right)-\frac{3{\left(-1\right)}^3+18{\left(-1\right)}^2+13\left(-1\right)-19}{9{\left(-1\right)}^2+36\left(-1\right)+13} \\ {x}_1=\left(-1\right)-\frac{-3+18-13-19}{9-36+13} \\ {x}_1=-1-\frac{17}{14} \\ {x}_1=-\frac{31}{14} \\ {x}_1=-2.21428 \end{gathered}$$

Similarly,

$$\begin{gathered} x_1=-2.21506708921454 \\ {x}_2=-1.86454803248149 \\ {x}_3=-1.86927525446249 \\ {x}_4=-1.86927382298528 \\ {x}_5=-1.86927382305909 \\ {x}_6=-1.86927382305908 \\ {x}_7=-1.86927382305908 \end{gathered}$$

Values are approximately equal.

The root is $m_2=-1.8692738$.

Assume, $x_0=-4$.

$$\begin{gathered} f\left(x_0\right)=3{x_0}^3+18{x_0}^2+13x_0-19 \\ f\text{'}\left(x_0\right)=9{x_0}^2+36x_0+13 \end{gathered}$$

Firstly, solve for $x_1$.

$$\begin{gathered} {\text{x}}_1=x_0-\frac{f\left(x_0\right)}{f\text{'}\left(x_0\right)} \\ {x}_1=\left(-4\right)-\frac{3{\left(-4\right)}^3+18{\left(-4\right)}^2+13\left(-4\right)-19}{9{\left(-4\right)}^2+36\left(-4\right)+13} \\ {x}_1=-4-\frac{25}{13} \\ {x}_1=-\frac{77}{13} \\ {x}_1=-5.92307 \end{gathered}$$

Similarly,

$$\begin{gathered} x_1=-5.92574289191213 \\ {x}_2=-5.16310903867916 \\ {x}_3=-4.87674689418742 \\ {x}_4=-4.83290545436777 \\ {x}_5=-4.8319222545269 \\ {x}_6=-4.83192226309298 \\ {x}_7=-4.83192226308854 \\ {x}_8=-4.83192226308854 \end{gathered}$$

Values are approximately equal.

The root is $m_3=-4.831922$.

The general solution$y=c_1{e}^{0.701t}+c_2{e}^{-1.869t}+c_3{e}^{-4.832t}$

Given:

The differential equation is$y^{iv}-5y\text{'}\text{'}+5y=0$

The auxiliary equation for the above equation$m^4-5m^2+5=0$

Similarly, using Newton’s method to find the roots of the above equation,

$m_1=-1.902,   m_2=-1.176,   m_3=1.176,   m_4=1.902$

The general solution$y=c_1{e}^{\left(-1.902\right)t}+c_2{e}^{\left(-1.176\right)t}+c_3{e}^{\left(1.176\right)t}+c_4{e}^{\left(1.902\right)t}$

Given:

The differential equation is$y^v-3y^{iv}-5y\text{'}\text{'}\text{'}+15y\text{'}\text{'}+4y\text{'}-12y=0$

The auxiliary equation for the above equation$m^5-3m^4-5m^3+15m^2+4m-12=0$

Similarly, using Newton’s method to find the roots of the above equation

$m_1=-2,   m_2=-1,   m_3=1,   m_4=2,   m_5=3$

The general solution

$y=c_1{e}^{\left(-2\right)t}+c_2{e}^{\left(-1\right)t}+c_3{e}^t+c_4{e}^{2t}+c_5{e}^{3t}$