Q12RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a general solution to the given differential equation. $\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{2t}\mathbf{+}{\mathbf{c}}_2{\mathbf{te}}^{2t}\mathbf{+}{\mathbf{c}}_3{\mathbf{e}}^{\frac{-5}{2}\mathbf{t}}$

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

The differential equation is $\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation $\mathbf{2}{\mathbf{m}}^3\mathbf{-}\mathbf{3}{\mathbf{m}}^2\mathbf{-}\mathbf{12}\mathbf{m}\mathbf{+}\mathbf{20}\mathbf{=}\mathbf{0}.$

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

Solve the auxiliary equation,

 $\begin{array}{c}\mathbf{2}{\mathbf{m}}^3\mathbf{-}\mathbf{3}{\mathbf{m}}^2\mathbf{-}\mathbf{12}\mathbf{m}\mathbf{+}\mathbf{20}\mathbf{=}\mathbf{0}\\ \mathbf{2}{\mathbf{m}}^3\mathbf{+}\mathbf{5}{\mathbf{m}}^2\mathbf{-}\mathbf{8}{\mathbf{m}}^2\mathbf{-}\mathbf{20}\mathbf{m}\mathbf{+}\mathbf{8}\mathbf{m}\mathbf{+}\mathbf{20}\mathbf{=}\mathbf{0}\\ {\mathbf{m}}^2\left(2m+5\right)\mathbf{-}\mathbf{4}\mathbf{m}\left(2m+5\right)\mathbf{+}\mathbf{4}\left(2m+5\right)\mathbf{=}\mathbf{0}\\ \left(2m+5\right)\left({\mathbf{m}}^2\mathbf{-}\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{4}\right)\mathbf{=}\mathbf{0}\\ \left(2m+5\right){\left(m-2\right)}^2\mathbf{=}\mathbf{0}\end{array}$

The roots of the auxiliary equation are, ${\mathbf{m}}_1\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{m}}_2\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{m}}_3\mathbf{=}\mathbf{-}\frac{5}{2}.$ 

Thus, the general solution of the given equation is $\mathbf{y}\mathbf{=}\left({\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{t}\right)e^{2t}\mathbf{+}{\mathbf{c}}_3{\mathbf{e}}^{\frac{-5}{2}\mathbf{t}}.$