Onsider the initial value problem y′′+7y′+12y=0, y(0)=6, y′(0)=β, where β>0. (b) Determine the coordinates tm and ymof the maximum point of the solution as functions of β. Enclose arguments of functions in parentheses. For example, sin(2x).

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) ).

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step-by-step explanation:

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