Answer:
[tex]y=e^{-3t}(A\: cos\: t+B\:sin\:t)[/tex]
Step-by-step explanation:
Given Second-Order Homogenous Differential Equation
[tex]y''+6y'+10y=0[/tex]
Use Auxiliary Equation
[tex]m^2+6m+10=0\\\\(m+3)^2+1=0\\\\(m+3)^2=-1\\\\m+3=\pm i\\\\m=-3\pm i[/tex]
General Solution
[tex]y=e^{pt}(A\: cos\: qt+B\:sin\:qt)\\\\y=e^{-3t}(A\: cos\: t+B\:sin\:t)[/tex]
Note that the DE has two distinct complex solutions [tex]p\pm qi[/tex] where [tex]A[/tex] and [tex]B[/tex] are arbitrary constants.
