Following are the calculation to the given function:
Given:
[tex]\bold{z''+8z'=0}[/tex]
To find:
real-valued=?
Solution:
Considering the differential equation that is: [tex]\bold{z''+8z'=0}[/tex]
When the auxiliary equation of the differential equation is:
[tex]\\ \bold{m^2 +8m =0}\\\\\bold{m(m+8) = 0 }\\\\\bold{m =0 \ \ \ \ \ \ \ m+8=0 }\\\\\bold{m =0 \ \ \ \ \ \ \ m=-8 }\\\\[/tex]
roots= [tex]\bold{0, \ -8}[/tex]
Calculating the general solution:
[tex]\bold{Z= c_1e^{0t}+C_2.e^{-8t}}\\\\ \bold{Z= c_1e+C_2.e^{-8t}}\\\\[/tex]
Learn more:
brainly.com/question/7508408
