Q1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$ 

Step-by-Step Solution

Answer

Thus, the general solution to the given differential equation is. $\mathbf{y}\mathbf{=}{\mathbf{C}}_1\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{-4x}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{2x}$

Step 1: Use the given equation to find a general solution for the differential equation with x

The given differential equation is,

 $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$

The auxiliary equation is.${\mathbf{m}}^3\mathbf{+}\mathbf{2}{\mathbf{m}}^2\mathbf{-}\mathbf{8}\mathbf{m}\mathbf{=}\mathbf{0}$

 

Find the roots of the auxiliary equation.

 $\begin{array}{c}{\mathbf{m}}^3\mathbf{+}\mathbf{2}{\mathbf{m}}^2\mathbf{-}\mathbf{8}\mathbf{m}\mathbf{=}\mathbf{0}\\ \mathbf{m}({\mathbf{m}}^2\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{-}\mathbf{8})\mathbf{=}\mathbf{0}\\ \mathbf{m}(\mathbf{m}\mathbf{+}\mathbf{4})(\mathbf{m}\mathbf{-}\mathbf{2})\mathbf{=}\mathbf{0}\\ m_1=0,m_2=-4,m_3=2\end{array}$

Step 2: General solution.

 

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

 

$\begin{array}{c}\mathbf{y}\mathbf{=}{\mathbf{C}}_1{\mathbf{e}}^{{\mathbf{m}}_1\mathbf{x}}\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{{\mathbf{m}}_2\mathbf{x}}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{{\mathbf{m}}_3\mathbf{x}}\\ \mathbf{y}\mathbf{=}{\mathbf{C}}_1{\mathbf{e}}^{\left(\mathbf{0}\right)\mathbf{x}}\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{(\mathbf{-}\mathbf{4})\mathbf{x}}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{\left(\mathbf{2}\right)\mathbf{x}}\\ \mathbf{y}\mathbf{=}{\mathbf{C}}_1\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{-4x}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{2x}\end{array}$

Thus, the general solution to the given differential equation is.$\mathbf{y}\mathbf{=}{\mathbf{C}}_1\mathbf{+}{\mathbf{C}}_2{\mathbf{e}}^{-4x}\mathbf{+}{\mathbf{C}}_3{\mathbf{e}}^{2x}$