Q19RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.

$\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is:

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

Write the auxiliary equation of the given differential equation

The given differential equation is, $\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

 

The auxiliary equation for the above equation $\mathbf{4}{\mathbf{m}}^{\mathbf{3}}\mathbf{+}\mathbf{8}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{11}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}$

Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{rcl}\mathbf{4}{\mathbf{m}}^{\mathbf{3}}\mathbf{+}\mathbf{8}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{11}\mathbf{m}\mathbf{+}\mathbf{3}& \mathbf{=}& \mathbf{0}\\ \mathbf{4}{\mathbf{m}}^{\mathbf{3}}\mathbf{+}\mathbf{12}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{12}\mathbf{m}\mathbf{+}\mathbf{m}\mathbf{+}\mathbf{3}& \mathbf{=}& \mathbf{0}\\ \mathbf{4}{\mathbf{m}}^{\mathbf{2}}\left(\mathbf{m}\mathbf{+}\mathbf{3}\right)\mathbf{-}\mathbf{4}\mathbf{m}\left(\mathbf{m}\mathbf{+}\mathbf{3}\right)\mathbf{+}\mathbf{1}\left(\mathbf{m}\mathbf{+}\mathbf{3}\right)& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{m}\mathbf{+}\mathbf{3}\right)\left(\mathbf{4}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{1}\right)& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{m}\mathbf{+}\mathbf{3}\right){\left(\mathbf{2}\mathbf{m}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}& \mathbf{=}& \mathbf{0}\\ & & \end{array}$

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

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