Q7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Answer

 

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

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

The given differential equation is $\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation $\mathbf{6}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}.$

 

Step 2: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

 $$\begin{gathered} \mathbf{6}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0} \\ \mathbf{6}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{m}\mathbf{-}\mathbf{3}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0} \\ \mathbf{2}\mathbf{m}\left(\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{2}\right)\mathbf{-}\mathbf{1}\left(\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{2}\right)\mathbf{=}\mathbf{0} \\ \left(\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{2}\right)\left(\mathbf{2}\mathbf{m}\mathbf{-}\mathbf{1}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

The roots of the auxiliary equation are

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}.$

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

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