Spring Differential Equation w/ Damping

[AmagicalFishy]

Homework Statement

A spring with a spring constant of 20 pounds per foot is loaded with a 10-pound weight and allowed to reach equilibrium. It is then displaced 1 foot downward and released. If the weight experiences a retarding force in pounds equal to four times the velocity at every point, find the equation of motion where t is time and y(t) is displacement with respect to time.

Homework Equations

<br /> 10y&#039;&#039; + 4y&#039; + 20y = 0 \\<br /> y(t) = C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t} \\<br /> e^{i\theta} = cos{\theta} + isin{\theta} \\<br /> y(0) = 1 \\<br /> y&#039;(0) = 0 \\<br />

The Attempt at a Solution

The ODE's characteristic equation is:
10\lambda ^2 + 4\lambda + 20 = 0 \\<br /> \text{or} \\<br /> 5\lambda ^2 + 2\lambda + 10 = 0
Solving for lambda...
\lambda = \frac{-1 \pm 7i}{5}

Therefore, y(t)—the solution to the ODE—is...
y(t) = C_1e^{\frac{-1 + 7i}{5} t} + C_2e^{\frac{-1 - 7i}{5} t} \\ <br /> \\<br /> = C_1e^{\frac{-1}{5}t}e^{\frac{7i}{5}t} + C_2e^{\frac{-1}{5}t}e^{\frac{-7i}{5}t} \\<br /> = e^{\frac{-1}{5}t} (C_1e^{\frac{7i}{5}t} + C_2e^{\frac{-7i}{5}t}) \\ <br /> = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{-7}{5}t} + C_2isin{\frac{-7}{5}t}) \\<br /> = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{-7}{5}t} + C_2isin{\frac{-7}{5}t})
Since cos[x] is even and sin[x] is odd...
<br /> = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{7}{5}t} - C_2isin{\frac{7}{5}t}) \\<br /> = e^{\frac{-1}{5}t} ( (C_1 + C_2)cos{\frac{7}{5}t} + (C_1 - C_2)isin{\frac{7}{5}t}) \\<br /> \text{Let:} \\<br /> A = (C_1 + C_2), B = i(C_1 - C_2) \\<br /> \text{So...} \\<br /> y(t) = e^{\frac{-1}{5}t} (Acos{\frac{7}{5}t} + Bsin{\frac{7}{5}t})<br />

Now we use the initial conditions to solve for A and B.

<br /> y(0) = e^{\frac{-1}{5}(0)} (A\cos{\frac{7}{5}(0)} + B\sin{\frac{7}{5}(0)}) = 1 \\<br /> = A\cos 0 + B\sin 0 = 1 \\<br /> A = 1 \\<br /> y&#039;(t) = \frac{-1}{5}e^{\frac{-1}{5}t}(A\cos{\frac{7}{5}t} + B\sin{\frac{7}{5}t}) +<br /> e^{\frac{-1}{5}t} (\frac{-7}{5}A\sin{\frac{7}{5}t} + \frac{7}{5}B\cos{\frac{7}{5}t})<br />

We end up w/: B = 1/7

So...
<br /> y(t) = e^{\frac{-1}{5}t} \left(\cos \frac{7}{5}t + \frac{1}{7}\sin \frac{7}{5}t \right)

(I'm not sure why the LaTeX stuff is showing up in boxes like that; I've tried a few things but it's always coming up).

This answer's, apparently, wrong; I'm not sure why. Halp, please.

 

[STEMucator]

$y(t) = c_1e^{\frac{-1 + 7i}{5} t} + c_2e^{\frac{-1 - 7i}{5} t} \\ = c_1e^{-\frac{t}{5}}(cos(\frac{7t}{5}) + isin(\frac{7t}{5})) + c_2e^{-\frac{t}{5}}(cos(\frac{7t}{5}) - isin(\frac{7t}{5}))$

Find y'.

Now use those initial conditions.

 

[vela]

Wolfram Alpha's answer agrees with yours. It could be you're entering the answer incorrectly if it's graded by a computer. I expect, however, it's because the coefficient of y'' in your differential equation is wrong. It's suppose to be the mass, not the weight of the object.