Answer:
Step-by-step explanation:
y′′+7y′+12y=0, y(0)=6, y′(0)=β, where β>0
This differential equation can be solved with the characteristic polinomyal:
m^(2) + 7m + 12m = 0
by using the quadratic formula: m = [-b ± sqrt(b^(2)-4*a*c)] / (2a)
we obtain:
m1 = -4
m2 = -3
Hence, the solution of the differential equation is
y(t) = A*e^(-4t) + B*e^(-6t)
Now, we have to use the initial conditions for the calculation of the constants A and B:
y(0) = A*e^(0) + B*e^(0) = A + B = 6
and the derivative
y'(t) = -4A*e^(-4t) - 6B*e^(-6t)
Y'(0) = -4A*e^(0) - 6B*e^(0) = -4A - 6B = β
Hence we have the system of equations
A + B = 6
-4A - 6B = β
By solving this system we have:
A = 18 + β/2
B = -12 - β/2
the maximum point is calculated by using y'(t) = 0
y'(t) = -4A*e^(-4t) - 6B*e^(-6t) = 0
with some algebra we can obtain that
e^(-2t)=-2A/3B --> Ln e^(-2t) = Ln (-2A/3B)
t = (-1/2)Ln(-2A/3B)
and by replacing we obtain the tmax
tmax = (-1/2)Ln( [2(36+ β)]/[3(24+ β)] "t value for a maximum"
an evaluating this t value in y(max) we can calculate the y value for the maximum coordinate P( tmax , y(tmax) ).
