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Expert-verified Found in: Page 332 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Find a general solution for the differential equation with x as the independent variable.${\mathbf{6}}{\mathbf{z}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{7}}{\mathbf{z}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{z}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{2}}{\mathbf{z}}{\mathbf{=}}{\mathbf{0}}$

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

$\mathbf{z}\mathbf{=}{\mathbf{C}}_{1}{\mathbf{e}}^{-x}\mathbf{+}{\mathbf{C}}_{2}{\mathbf{e}}^{\frac{x}{2}}\mathbf{+}{\mathbf{C}}_{3}{\mathbf{e}}^{\frac{-2x}{3}}$
See the step by step solution

## 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.

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

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.

$\begin{array}{c}\mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{±}\sqrt{\mathbf{1}\mathbf{+}\mathbf{4}\left(\mathbf{12}\right)}}{12}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{±}\sqrt{49}}{12}\\ \mathbf{m}\mathbf{=}\frac{-1±7}{12}\\ \mathbf{m}\mathbf{=}\frac{-1+7}{12}\mathbf{,}\text{\hspace{0.17em}}\frac{-1-7}{12}\\ {\mathbf{m}}_{2}\mathbf{=}\frac{1}{2}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{m}}_{3}\mathbf{=}\mathbf{-}\frac{2}{3}\end{array}$

## Step 2: Write the general solution.

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

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

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

$\mathbf{z}\mathbf{=}{\mathbf{C}}_{1}{\mathbf{e}}^{-x}\mathbf{+}{\mathbf{C}}_{2}{\mathbf{e}}^{\frac{x}{2}}\mathbf{+}{\mathbf{C}}_{3}{\mathbf{e}}^{\frac{-2x}{3}}$ ### Want to see more solutions like these? 