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**Ch 3 / Re: Problem of the week 4a**

« **on:**January 31, 2013, 07:25:00 PM »

Note that the greater of the two roots in case 1 is always greater than the root $-\alpha/2 = -\omega$ in case 2. We call case 1

The solution to equation (3) is given by the sum of the general solution to the homogeneous equation (1) and a particular solution to (3) which we will solve by the method of undetermined coefficients. As hinted, let $y_p = A \cos (\beta t) + B \sin(\beta t)$. Then

\begin{align}

y'' + \alpha y' + \omega^2 y

&= (A \cos(\beta t) + B \sin(\beta t))'' + \alpha (A \cos(\beta t) + B \sin(\beta t))' + \omega^2 (A \cos (\beta t) + B \sin(\beta t)) \\

&= -A\beta^2 \cos(\beta t) - B\beta^2 \sin(\beta t) + -A\alpha\beta\sin(\beta t) + B\alpha\beta\cos(\beta t) + A\omega^2 \cos(\beta t) + B\omega^2 \sin(\beta t) \\

&= (-A\beta^2 + B\alpha\beta + A\omega^2)\cos(\beta t) + (-B\beta^2 -A\alpha\beta + B\omega^2)\sin(\beta t)

\end{align}

By matching coefficients we find

\begin{align}

(-\beta^2 + \omega^2) A + \alpha\beta B &= 1 \\

(-\alpha\beta) A + (-\beta^2 + \omega^2) B &= 0

\end{align}

This is a linear system in two equations and two unknowns, which is therefore easy to solve. The solution is

\begin{align}

A &= \frac{\omega^2 - \beta^2}{(\omega^2-\beta^2)^2 + (\alpha \beta)^2} \\

B &= \frac{\alpha\beta}{(\omega^2 -\beta^2)^2 + (\alpha \beta)^2}

\end{align}

so the particular solution to (3) is $y_p = A \cos(\beta t) + B\sin(\beta t)$ with $A, B$ given above and the general solution to (3) is given by adding $y_p$ and the general solution to (1) as found above.

(continued)

*overdamped*because $\alpha$ is large and the greater root (i.e., less negative) leads to a slower exponential decay. In contrast, case 2 is called*critically damped*because solutions decay with time constant $\alpha/2 = \omega$, which is maximal. Case 3 is called*underdamped*because $\alpha$ is small and the damping force is insufficient to prevent oscillation. In this case the decay will also be slow since the time constant is $\alpha/2 < \omega$.The solution to equation (3) is given by the sum of the general solution to the homogeneous equation (1) and a particular solution to (3) which we will solve by the method of undetermined coefficients. As hinted, let $y_p = A \cos (\beta t) + B \sin(\beta t)$. Then

\begin{align}

y'' + \alpha y' + \omega^2 y

&= (A \cos(\beta t) + B \sin(\beta t))'' + \alpha (A \cos(\beta t) + B \sin(\beta t))' + \omega^2 (A \cos (\beta t) + B \sin(\beta t)) \\

&= -A\beta^2 \cos(\beta t) - B\beta^2 \sin(\beta t) + -A\alpha\beta\sin(\beta t) + B\alpha\beta\cos(\beta t) + A\omega^2 \cos(\beta t) + B\omega^2 \sin(\beta t) \\

&= (-A\beta^2 + B\alpha\beta + A\omega^2)\cos(\beta t) + (-B\beta^2 -A\alpha\beta + B\omega^2)\sin(\beta t)

\end{align}

By matching coefficients we find

\begin{align}

(-\beta^2 + \omega^2) A + \alpha\beta B &= 1 \\

(-\alpha\beta) A + (-\beta^2 + \omega^2) B &= 0

\end{align}

This is a linear system in two equations and two unknowns, which is therefore easy to solve. The solution is

\begin{align}

A &= \frac{\omega^2 - \beta^2}{(\omega^2-\beta^2)^2 + (\alpha \beta)^2} \\

B &= \frac{\alpha\beta}{(\omega^2 -\beta^2)^2 + (\alpha \beta)^2}

\end{align}

so the particular solution to (3) is $y_p = A \cos(\beta t) + B\sin(\beta t)$ with $A, B$ given above and the general solution to (3) is given by adding $y_p$ and the general solution to (1) as found above.

(continued)