Answer:
[tex]y(t) = C_1e^{4t} + C_2e^{-7t}[/tex]
Step-by-step explanation:
We are given the following information in the question:
[tex]y''+3y'-28y = 0[/tex]
This can be written as:
[tex](D^2 + 3D - 28)y = 0[/tex]
The auxiliary equation obtained is:
[tex]m^2+ 3m - 28 = 0\\ m^2 + 7m - 4m - 28 = 0\\m(m+7)-4(m+7)=0\\(m-4)(m+7)\\m = 4, m = -7[/tex]
Thus, the general solution of the equation is:
[tex]y(t) = C_1e^{4t} + C_2e^{-7t}\\\text{where } C_1, C_2 \text{ are constants}[/tex]
