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

[tex]
10y'' + 4y' + 20y = 0 \\
y(t) = C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t} \\
e^{i\theta} = cos{\theta} + isin{\theta} \\
y(0) = 1 \\
y'(0) = 0 \\
[/tex]

The Attempt at a Solution

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

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

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

[tex]
y(0) = e^{\frac{-1}{5}(0)} (A\cos{\frac{7}{5}(0)} + B\sin{\frac{7}{5}(0)}) = 1 \\
= A\cos 0 + B\sin 0 = 1 \\
A = 1 \\
y'(t) = \frac{-1}{5}e^{\frac{-1}{5}t}(A\cos{\frac{7}{5}t} + B\sin{\frac{7}{5}t}) +
e^{\frac{-1}{5}t} (\frac{-7}{5}A\sin{\frac{7}{5}t} + \frac{7}{5}B\cos{\frac{7}{5}t})
[/tex]

We end up w/: B = 1/7

So...
[tex]
y(t) = e^{\frac{-1}{5}t} \left(\cos \frac{7}{5}t + \frac{1}{7}\sin \frac{7}{5}t \right)[/tex]

(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.