Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Answer

 

The general solution of the given equation is

 $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}.$

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

Solve the auxiliary equation,

 

$$\begin{gathered} {\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathbf{m}\mathbf{+}\mathbf{6}\mathbf{=}\mathbf{0} \\ {\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{6}\mathbf{=}\mathbf{0} \\ \mathbf{m}\left(\mathbf{m}\mathbf{-}\mathbf{3}\right)\mathbf{-}\mathbf{2}\left(\mathbf{m}\mathbf{-}\mathbf{3}\right)\mathbf{=}\mathbf{0} \\ \left(\mathbf{m}\mathbf{-}\mathbf{2}\right)\left(\mathbf{m}\mathbf{-}\mathbf{3}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

 The roots of the auxiliary equation are ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{3}.$

 

Step 3: Write the general solution.

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

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

Thus, the general solution of the given equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}.$