Q1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a general solution to the given differential equation.

$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Answer

 

The general solution of the given equation is ${\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{t}}.}_{}$

 

Step 1: Write the auxiliary equation of the given differential equation.

The given differential equation is $\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}.$$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation is $\mathbf{2}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathbf{m}\mathbf{-}\mathbf{4}\mathbf{=}\mathbf{0}.$

Step 2: Find the roots of the auxiliary equation.

Solve the auxiliary equation,

 $$\begin{gathered} \mathbf{2}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathbf{m}\mathbf{-}\mathbf{4}\mathbf{=}\mathbf{0} \\ \mathbf{2}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{8}\mathbf{m}\mathbf{-}\mathbf{m}\mathbf{-}\mathbf{4}\mathbf{=}\mathbf{0} \\ \mathbf{2}\mathbf{m}\left(\mathbf{m}\mathbf{+}\mathbf{4}\right)\mathbf{-}\mathbf{1}\left(\mathbf{m}\mathbf{+}\mathbf{4}\right)\mathbf{=}\mathbf{0} \\ \left(\mathbf{m}\mathbf{+}\mathbf{4}\right)\left(\mathbf{2}\mathbf{m}\mathbf{-}\mathbf{1}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

The roots of the auxiliary equation are ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{4}.$

 

When an auxiliary equation has distinct real roots as&, then the general solution is given as;

 $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{{\mathbf{m}}_{\mathbf{1}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{{\mathbf{m}}_{\mathbf{2}}\mathbf{t}}$

 the general solution of the given equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{t}}.$