Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: find a general solution to the given differential equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{=}\mathbf{0}$.

Step-by-Step Solution

Answer

Answer

 

The general solution of the given equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{1}\mathbf{+}\mathbf{3}\sqrt{\mathbf{5}}}{\mathbf{2}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\frac{\mathbf{1}\mathbf{-}\mathbf{3}\sqrt{\mathbf{5}}}{\mathbf{2}}\mathbf{t}}.$

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

The given differential equation is $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation ${\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{m}\mathbf{-}\mathbf{11}\mathbf{=}\mathbf{0}.$

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

Solve the auxiliary equation,

$$\begin{gathered} {\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{m}\mathbf{-}\mathbf{11}\mathbf{=}\mathbf{0} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\left(\mathbf{-}\mathbf{1}\right)\mathbf{\pm }\sqrt{\mathbf{1}\mathbf{-}\mathbf{4}\left(\mathbf{1}\right)\left(\mathbf{-}\mathbf{11}\right)}}{\mathbf{2}} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{1}\mathbf{\pm }\sqrt{\mathbf{45}}}{\mathbf{2}} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{1}\mathbf{\pm }\mathbf{3}\sqrt{\mathbf{5}}}{\mathbf{2}} \end{gathered}$$

The roots of the auxiliary equation are

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{3}\sqrt{\mathbf{5}}}{\mathbf{2}}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}\mathbf{-}\mathbf{3}\sqrt{\mathbf{5}}}{\mathbf{2}}.$

Step 3: Write the general solution.

If an auxiliary equation has distinct real roots&, 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 

$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{1}\mathbf{+}\mathbf{3}\sqrt{\mathbf{5}}}{\mathbf{2}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\frac{\mathbf{1}\mathbf{-}\mathbf{3}\sqrt{\mathbf{5}}}{\mathbf{2}}\mathbf{t}}.$