Q9RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.$\mathbf{16}\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{56}\mathbf{z}\mathbf{\text{'}}\mathbf{+}\mathbf{49}\mathbf{z}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Thus, the general solution to the given differential equation is$\mathbf{z}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{7}{4}\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{te}}^{\frac{7}{4}\mathbf{t}}.$

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

The differential equation is$\mathbf{16}\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{56}\mathbf{z}\mathbf{\text{'}}\mathbf{+}\mathbf{49}\mathbf{z}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation$\mathbf{16}{\mathbf{m}}^2\mathbf{-}\mathbf{56}\mathbf{m}\mathbf{+}\mathbf{49}\mathbf{=}\mathbf{0}.$

Step 2: Find the roots of the auxiliary equation.

Solve the auxiliary equation,

 $\begin{array}{c}\mathbf{16}{\mathbf{m}}^2\mathbf{-}\mathbf{56}\mathbf{m}\mathbf{+}\mathbf{49}\mathbf{=}\mathbf{0}\\ {\left(4m-7\right)}^2\mathbf{=}\mathbf{0}\end{array}$

The roots of the auxiliary equation are${\mathbf{m}}_1\mathbf{=}\frac{7}{4}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\frac{7}{4}.$

 

The general solution of the given equation is $\mathbf{z}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{7}{4}\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{te}}^{\frac{7}{4}\mathbf{t}}.$