Respuesta :
Answer:
k = 30, [tex]y(t) = C_1e^{5t}+C_2e^{6t}[/tex]
Step-by-step explanation:
Since [tex]y=e^{5t}[/tex] is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that
[tex]\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}[/tex]
Then, replacing the derivatives in the equation we have:
[tex]25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0[/tex]
Since [tex]e^{5t}[/tex] is a positive function, we have that
[tex]25-55+k = 0 \rightarrow k = 30[/tex].
Now, consider a general solution [tex]y(t) = Ae^{rt}, A \in \mathbb{R}[/tex], then, by calculating the derivatives and replacing them in the equation, we get
[tex]Ae^{rt}(r^2-11r+30)=0[/tex]
We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.
Therefore, the general solution is
[tex]y(t) = C_1e^{5t}+C_2e^{6t}[/tex]
