Answer:
[tex]y(t)=7e^{-3t}+3e^{-5t}[/tex]
Step-by-step explanation:
Solve the given initial-value problem.
[tex]y" + 8y + 15y = 0; \ y(0) = 10, \ y'(0) = -36[/tex]
(1) - Form and solve the characteristic equation for "m"
[tex]y" + 8y + 15y = 0 \\\\\\\Longrightarrow \boxed{m^2+8m+15}\\\\\\m^2+8m+15\\\\\\\Longrightarrow (m+3)(m+5)=0\\\\\\\therefore m=-3, \ -5[/tex]
(2) - Form the general solution
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Solutions to Higher-order DE's:}}\\\\\text{Real,distinct roots} \rightarrow y=c_1e^{m_1t}+c_2e^{m_2t}+...+c_ne^{m_nt}\\\\ \text{Duplicate roots} \rightarrow y=c_1e^{mt}+c_2te^{mt}+...+c_nt^ne^{mt}\\\\ \text{Complex roots} \rightarrow y=c_1e^{\alpha t}\cos(\beta t)+c_2e^{\alpha t}\sin(\beta t)+... \ ;m=\alpha \pm \beta i\end{array}\right}[/tex]
The roots are real and distinct, so we can form the general solution as:
[tex]\therefore y=c_1e^{-3t}+c_2e^{-5t}[/tex]
(3) - Now use the given initial conditions to determine the values for the arbitrary constants c_1 and c_2
[tex]y=c_1e^{-3t}+c_2e^{-5t}; \ \text{Recall} \rightarrow \ y(0) = 10, \ y'(0) = -36\\\\\\y=c_1e^{-3t}+c_2e^{-5t}\\\\y'=-3c_1e^{-3t}-5c_2e^{-5t}\\\\\text{Plugging in the initial conditions:}\\ \Longrightarrow \left\{\begin{array}{cc}10=c_1+c_2\\-36=-3c_1-5c_2\end{array}\right\\\\\\\text{After solving the system we get:} \ \boxed{c_1=7, \ c_2=3}[/tex]
(4) - Now we can form the solution
[tex]\therefore \boxed{\boxed{y(t)=7e^{-3t}+3e^{-5t}}}[/tex]
