Q3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for the differential equation with x as the independent variable.$\mathbf{6}\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{z}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Thus, the general solution to the given differential equation is;

 

$\mathbf{z}\mathbf{=}{\mathbf{C}}_1{\mathbf{e}}^{-x}\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{\frac{x}{2}}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{\frac{-2x}{3}}$

Step 1: Write the auxiliary equation.

The given differential equation is;

 $\mathbf{6}\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{z}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{=}\mathbf{0}$ 

The auxiliary equation is .$\mathbf{6}{\mathbf{m}}^3\mathbf{+}\mathbf{7}{\mathbf{m}}^2\mathbf{-}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}$

 

Simplify the auxiliary equation.

 $\begin{array}{c}\mathbf{6}{\mathbf{m}}^3\mathbf{+}\mathbf{7}{\mathbf{m}}^2\mathbf{-}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}\\ \mathbf{6}{\mathbf{m}}^3\mathbf{+}\mathbf{7}{\mathbf{m}}^2\mathbf{-}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}\\ (\mathbf{m}\mathbf{+}\mathbf{1})(\mathbf{6}{\mathbf{m}}^2\mathbf{+}\mathbf{m}\mathbf{-}\mathbf{2})\mathbf{=}\mathbf{0}\end{array}$

One of the roots is . ${\mathbf{m}}_1\mathbf{=}\mathbf{-}\mathbf{1}$

Find the other roots of the auxiliary equation by solving the quadratic equation.

 $\begin{array}{c}\mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{\pm }\sqrt{\mathbf{1}\mathbf{+}\mathbf{4}\left(\mathbf{12}\right)}}{12}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{\pm }\sqrt{49}}{12}\\ \mathbf{m}\mathbf{=}\frac{-1\pm 7}{12}\\ \mathbf{m}\mathbf{=}\frac{-1+7}{12}\mathbf{,}\frac{-1-7}{12}\\ {\mathbf{m}}_2\mathbf{=}\frac{1}{2}\mathbf{,}{\mathbf{m}}_3\mathbf{=}\mathbf{-}\frac{2}{3}\end{array}$

Step 2: Write the general solution.

The roots are real and distinct; therefore the general solution to the given differential equation is given as:

 $\begin{array}{c}\mathbf{z}\mathbf{=}{\mathbf{C}}_1{\mathbf{e}}^{{\mathbf{m}}_1\mathbf{x}}\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{{\mathbf{m}}_2\mathbf{x}}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{{\mathbf{m}}_3\mathbf{x}}\\ \mathbf{z}\mathbf{=}{\mathbf{C}}_1{\mathbf{e}}^{(\mathbf{-}\mathbf{1})\mathbf{x}}\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{\left(\frac{1}{2}\right)\mathbf{x}}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{(\mathbf{-}\frac{2}{3})\mathbf{x}}\\ \mathbf{z}\mathbf{=}{\mathbf{C}}_1{\mathbf{e}}^{-x}\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{\frac{x}{2}}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{\frac{-2x}{3}}\end{array}$

Thus, the general solution to the given differential equation is;

 $\mathbf{z}\mathbf{=}{\mathbf{C}}_1{\mathbf{e}}^{-x}\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{\frac{x}{2}}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{\frac{-2x}{3}}$