Q12E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{11}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Answer

 

The general solution of the given equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{11}\mathbf{+}\sqrt{\mathbf{205}}}{\mathbf{6}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{11}\mathbf{-}\sqrt{\mathbf{205}}}{\mathbf{6}}\mathbf{t}}.$

Step 1: Firstly, write the auxiliary equation of the given differential equation.

The given differential equation is $\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{11}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation $\mathbf{3}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{11}\mathbf{m}\mathbf{-}\mathbf{7}\mathbf{=}\mathbf{0}.$

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

Solve the auxiliary equation,

$$\begin{gathered} \mathbf{3}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{11}\mathbf{m}\mathbf{-}\mathbf{7}\mathbf{=}\mathbf{0} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{11}\mathbf{\pm }\sqrt{{\left(\mathbf{11}\right)}^{\mathbf{2}}\mathbf{-}\mathbf{4}\left(\mathbf{3}\right)\left(\mathbf{-}\mathbf{7}\right)}}{\mathbf{2}\left(\mathbf{3}\right)} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{11}\mathbf{\pm }\sqrt{\mathbf{121}\mathbf{+}\mathbf{84}}}{\mathbf{6}} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{11}\mathbf{\pm }\sqrt{\mathbf{205}}}{\mathbf{6}} \end{gathered}$$ 

The roots of the auxiliary equation are

 ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{-}\mathbf{11}\mathbf{+}\sqrt{\mathbf{205}}}{\mathbf{6}}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{-}\mathbf{11}\mathbf{-}\sqrt{\mathbf{205}}}{\mathbf{6}}.$

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 $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{11}\mathbf{+}\sqrt{\mathbf{205}}}{\mathbf{6}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{11}\mathbf{-}\sqrt{\mathbf{205}}}{\mathbf{6}}\mathbf{t}}.$