Q45E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

By using Newton’s method or some other numerical procedure to approximate the roots of the auxiliary equation, find general solutions to the following equations.

$$\begin{gathered} \left(\mathit{a}\right) \mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{18}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{13}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{19}\mathit{y}\mathbf{=}\mathbf{0} \\ \left(\mathit{b}\right) {\mathit{y}}^{\mathit{i}\mathit{v}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{0} \\ \left(\mathit{c}\right) {\mathit{y}}^{\mathit{v}}\mathbf{-}\mathbf{3}{\mathit{y}}^{\mathit{i}\mathit{v}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{15}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathit{y}\mathbf{=}\mathbf{0} \end{gathered}$$

Step-by-Step Solution

Answer

a. $y=c_1{e}^{0.701t}+c_2{e}^{-1.869t}+c_3{e}^{-4.832t}$

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

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

Step 1: Firstly, write the auxiliary equation of the given differential equation

Given:

 

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

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

Step 2: Now use Newton’s method

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 $m_1=0.701190$.

Step 3: Now assume x 0 = - 1 and find the second root

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 $m_2=-1.8692738$. 

Step 4: Now assume x 0 = - 4 and find the third root

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

Step 5: Now use Newton’s method for part (b) and find the general solution

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

Step 6: Now use Newton’s method for part (c) and find a general solution.

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

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