Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for the differential equation with x as the independent variable.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Thus, the general solution to the given differential equation is; $\mathbf{y}\mathbf{=}{\mathbf{C}}_1{\mathbf{e}}^{3x}\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^x\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{-x}$

Step 1: Find the auxiliary equation

The given differential equation is;

 $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation is,

 ${\mathbf{m}}^3\mathbf{-}\mathbf{3}{\mathbf{m}}^2\mathbf{-}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}$

Take a fraction of the above equation,

$\begin{array}{c}{\mathbf{m}}^2(\mathbf{m}\mathbf{-}\mathbf{3})\mathbf{-}\mathbf{1}(\mathbf{m}\mathbf{-}\mathbf{3})\mathbf{=}\mathbf{0}\\ (\mathbf{m}\mathbf{-}\mathbf{3})({\mathbf{m}}^2\mathbf{-}\mathbf{1})\mathbf{=}\mathbf{0}\\ {\mathbf{m}}_1\mathbf{=}\mathbf{3}\mathbf{,}{\mathbf{m}}_2\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{m}}_3\mathbf{=}\mathbf{-}\mathbf{1}\end{array}$

Step 2: Write the general solution

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

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

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

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