Question

Find a general solution to the given differential equation.$\mathbf{36}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{24}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}$

Short answer

The general solution to the given differential equation is:

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{-1}{3}\mathbf{t}}\mathbf{cos}\left(\frac{t}{6}\right)\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{-1}{3}\mathbf{t}}\mathbf{sin}\left(\frac{t}{6}\right)$

Step-by-step solution

Complex conjugate roots.

If the auxiliary equation has complex conjugate roots, then the general solution is given as:

$y\left(t\right)=c_1{e}^{\alpha t}cos\beta t+c_2{e}^{\alpha t}sin\beta t$

Write the auxiliary equation of the given differential equation

The differential equation is$\mathbf{36}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{24}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}.$

The auxiliary equation for the above equation$\mathbf{36}{\mathbf{m}}^2\mathbf{+}\mathbf{24}\mathbf{m}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0}.$

Find the roots of the auxiliary equation. 

Solve the auxiliary equation,

$\begin{array}{c}\mathbf{36}{\mathbf{m}}^2\mathbf{+}\mathbf{24}\mathbf{m}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{24}\mathbf{\pm }\sqrt{576-720}}{\mathbf{2}\left(36\right)}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{24}\mathbf{\pm }\sqrt{-144}}{72}\\ \mathbf{m}\mathbf{=}\frac{-24\pm 12i}{72}\\ \mathbf{m}\mathbf{=}\frac{-1}{3}\mathbf{\pm }\frac{i}{6}\end{array}$

The roots of the auxiliary equation are,${\mathbf{m}}_1\mathbf{=}\frac{-1}{3}\mathbf{+}\frac{i}{6}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\frac{-1}{3}\mathbf{-}\frac{i}{6}.$

Thus, the general solution of the given equation is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{-1}{3}\mathbf{t}}\mathbf{cos}\left(\frac{t}{6}\right)\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{-1}{3}\mathbf{t}}\mathbf{sin}\left(\frac{t}{6}\right).$