If a, b, and c are positive constants, show that all solutions of ay\" + by\' +
ID: 3108724 • Letter: I
Question
If a, b, and c are positive constants, show that all solutions of ay" + by' + cy = 0 approach 0 as t rightarrow infinity. Since this equation models many electrical and mechanical networks, this important result states that there is no "perpetual motion". If a > 0 and c > 0 but b = 0, show that the results of Problem 5.1 are no longer true, but that all solutions are bounded as t rightarrow infinity. If a > 0 and b > 0 but c = 0, show that the results of Problem 5.1 are no longer true, but that all solutions approach a constant that depends on the initial conditions as t rightarrow infinity. Determine this constant for the initial conditions y(0) = y0, y'(0) = V0. Now consider the equation ay" + by' + cy = g(t). where a, b, and c are positive constants. If y_1(t) and Y_2(t) are two solutions of this non-homogeneous equation, show that Y_1 (t) - Y_2(t) rightarrow 0 as t rightarrow infinity. Suppose that g(t) = d is also a positive constant. Show that every solution approaches the constant d/c as t rightarrow infinity. Suppose that g(t) = d > 0, and that a > 0 and b > 0, but c = 0. What happens to the solutions as t rightarrow infinity?Explanation / Answer
Solved the first two problems, post multiple question to get the remaining answers
Problem 5.1
ay'' + by' + cy = 0
writing the characteristic equation of the homogeneous differential equation
ap^2 + bp + c = 0
D = b^2 - 4ac
p1 = [-b + sqrt(D)]/2a and p2 = [-b-sqrt(D)]/2a
Hence the solution of the differential equation will be
y(t) = c1e^(p1t) + c2e^(p2t)
Since all a,b and c are positive, hence p1 and p2 wil have negative real part, hence e^(p1t) will be tending to e^(-infinity) which will tend to zero
Hence both the solutions of the equation will tend to zero
Therefore, when t tends to infinity both the solutions of the differential equation will boil down to zero
Problem 5.2
ay'' + by' + cy = 0
writing the characteristic equation of the homogeneous differential equation
ap^2 + bp + c = 0
D = 0^2 - 4ac = -4ac
p1 = [sqrt(D)]/2a and p2 = [-sqrt(D)]/2a
Hence the solution of the differential equation will be
y(t) = c1e^(p1t) + c2e^(p2t)
Since the value of b is equal to zero, hence D is negative therefore the value of sqrt(D) will be negative hence the solution of the equation will be of the form
y(t) = c1cos(sqrt(D)t) + c2sin(sqrt(D)t)
Since cos and sin are bounded functions, hence the solution will be bounded when t tends to infinity
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