Applied Math (Biological application)
posted by Joey .
The deformation y(t) of heart muscle from its rest position after t seconds can be modelled as a springmass system, described by the differential equation:
my" + vy' + ky = F(t)
where m, v, and k are positive constants denoting the mass of the myocardial tissue, and the stiffness and damping coefficients, respectively; F(t) is called a forcing function.
1. In case of zero and assuming v = sqrt(4mk) ; find the general solution for the displacement of the heart tissue at time t > 0 (your answer will contain m, k, t, and arbritrary constants.)

Alrighty so this is how I chose to attempt the question:
1. First I converted the equation to auxillary form, meaning the y" became r^2, y' became r, and y became 1. This gives me mr^2 + vr + k. I make the righthand side = to 0 and solve using the quadratic formula. What I get is r = (sqrt(mk))/(m). This yields the solution y=e^((sqrt(mk))/(m)x).
I am having to do an alt. method for this question because it is considered an example of "repeated roots" (look at outcome of quadratic equation) so I am now having to do a full out different process from usual. Can anybody help out here? Thanks.

I basically need help finding the rest of the solutions.
Respond to this Question
Similar Questions

math
Suppose that an object attached to a coiled spring is pulled down a distance of 5 inches from its rest position and then released. If the time for one oscillation is 3 seconds, write and equation that relates the displacement d of … 
physics
A mass of 0.755 kg is attached to a horizontal spring of constant 403 N/m as shown. The system is compressed a distance xa = 20.8 cm from its equilibrium position and released from rest. The coefficient of friction on the surface is … 
calculus
Question 1: Experimentally it is found that a 6 kg weight stretches a certain spring 6 cm. If the weight is pulled 4 cm below the equilibrium position and released: a. Set up the differential equation and associated conditions describing … 
differential equations
Search: A 8 kg mass is attached to a spring hanging from the ceiling and allowed to come to rest. assume that the spring constant is 40 n/ m and a damping constant is 3 N/m. At t=0 and external force of 2 sin(2t+PI/4)N is applied to … 
physics
A mass of 0.7 kg is attached to the end of a massless spring of spring constant 0.35 N/m. It is released from rest from an extended position. After 0.8 s, the speed of the mass is measured to be 1.5 m/s. What is the amplitude of oscillation? 
Math
A frictionless spring with a 7kg mass can be held stretched 1 meters beyond its natural length by a force of 20 newtons. If the spring begins at its springmass equilibrium position, but a push gives it an initial velocity of 0.5 … 
physics
When a 1.3kg mass was attached to a spring, the spring stretched 21cm from its equilibrium position. The mass spring system was then set into oscillatory motion by providing the mass with an initial velocity. Its oscillatory motion … 
Help with differential eqs problem???? (Calculus)
Consider the differential equation dy/dt=yt a) Determine whether the following functions are solutions to the given differential equation. y(t) = t + 1 + 2e^t y(t) = t + 1 y(t) = t + 2 b) When you weigh bananas in a scale at the grocery … 
Health
3. The muscular system allows our bones to: A. Strenthen B. Contract C. Lenghthen D. Move***** 4. This muscle helps to transport blood to and from the heart: A. Smooth Muscle***** B. cardiac Muscle C. Skeletal Muscle D. Invoulntary … 
math
A spring is stretched 5 inches abobe its rest position. Itis then released and oscillates in damped harmontic motin with a frequency of 2cycles per second. After 7 seconds you measure the amplitude of the spring to be 1.5cm. 1.Find …