Q8RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.$\mathbf{25}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{-2}{5}\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{te}}^{\frac{-2}{5}\mathbf{t}}.$

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

The given differential equation is$\mathbf{25}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation$\mathbf{25}{\mathbf{m}}^2\mathbf{+}\mathbf{20}\mathbf{m}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}.$

Step 2: Find the roots of the auxiliary equation.

Solve the auxiliary equation, 

$\begin{array}{c}\mathbf{25}{\mathbf{m}}^2\mathbf{+}\mathbf{20}\mathbf{m}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}\\ {\left(5m+2\right)}^2\mathbf{=}\mathbf{0}\end{array}$

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

 

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