Q5RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation. $\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{3}{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{1}{3}\mathbf{t}}.$

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

The differential equation is$\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation is$\mathbf{6}{\mathbf{m}}^2\mathbf{-}\mathbf{11}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}.$

Step 2: Find the roots of the auxiliary equation

Solve the auxiliary equation,

 $\begin{array}{c}\mathbf{6}{\mathbf{m}}^2\mathbf{-}\mathbf{11}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}\\ \mathbf{6}{\mathbf{m}}^2\mathbf{-}\mathbf{9}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}\\ \mathbf{3}\mathbf{m}\left(2m-3\right)\mathbf{-}\mathbf{1}\left(2m-3\right)\mathbf{=}\mathbf{0}\\ \left(2m-3\right)\left(3m-1\right)\mathbf{=}\mathbf{0}\end{array}$

 

The roots of the auxiliary equation are, ${\mathbf{m}}_1\mathbf{=}\frac{3}{2}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\frac{1}{3}.$

Thus, the general solution of the given equation is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{3}{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{1}{3}\mathbf{t}}.$