Question

Let \(\delta(t)\) be the unit impulse function defined by \(\delta(x-x_0)=0\) when \(x\ne x_0\) and \(\displaystyle\int_{-\infty}^{\infty}\delta(x-x_0)\,dx=1\).
Consider the following initial value problem \[ 2\frac{d^2y}{dx^2}+\frac{dy}{dx}+2y=\delta(x-5) \] with \(y(0)=0\) and \(\dfrac{dy}{dx}(0)=0\). Which one of the following is TRUE?

• A. \(y(10)=\dfrac{10}{\sqrt{15}}e^{-5/4}\sin\left(\dfrac{\sqrt{15}}{4}\right)\)
• B. \(y(10)=\dfrac{2}{\sqrt{15}}e^{-5/4}\sin\left(\dfrac{5\sqrt{15}}{4}\right)\)
• C. \(y(10)=\dfrac{10}{\sqrt{15}}e^{5/4}\sin\left(\dfrac{\sqrt{15}}{4}\right)\)
• D. \(y(10)=\dfrac{2}{\sqrt{15}}e^{5/4}\sin\left(\dfrac{5\sqrt{15}}{4}\right)\)

Hint:

Take the Laplace transform of the ODE, use \(\mathcal{L}\{\delta(x-5)\}=e^{-5s}\), and invert with the time-shift property.

Answer 1

The Correct Option is B

Step 1: Take the Laplace transform of the whole equation.
Using \(\mathcal{L}\{y''\}=s^2Y(s)-sy(0)-y'(0)\) and \(\mathcal{L}\{y'\}=sY(s)-y(0)\), and since \(y(0)=0\) and \(y'(0)=0\), these reduce to \(\mathcal{L}\{y''\}=s^2Y(s)\) and \(\mathcal{L}\{y'\}=sY(s)\). Also \(\mathcal{L}\{\delta(x-5)\}=e^{-5s}\).
So the transformed equation is \(2s^2Y(s)+sY(s)+2Y(s)=e^{-5s}\), which gives
\[ Y(s) = \frac{e^{-5s}}{2s^2+s+2} \]
Step 2: Rewrite the denominator by completing the square.
\[ 2s^2+s+2 = 2\left[\left(s+\frac{1}{4}\right)^2 + \frac{15}{16}\right] \] so \(\dfrac{1}{2s^2+s+2} = \dfrac{1}{2}\cdot\dfrac{1}{(s+\frac14)^2+\frac{15}{16}}\).

Step 3: Invert the Laplace transform.
Using \(\mathcal{L}^{-1}\left\{\dfrac{1}{(s+a)^2+w^2}\right\} = \dfrac{1}{w}e^{-at}\sin(wt)\) with \(a=\frac14\) and \(w=\dfrac{\sqrt{15}}{4}\), the inverse transform of \(\dfrac{1}{2s^2+s+2}\) is
\[ g(t) = \frac{1}{2}\cdot\frac{4}{\sqrt{15}}e^{-t/4}\sin\left(\frac{\sqrt{15}}{4}t\right) = \frac{2}{\sqrt{15}}e^{-t/4}\sin\left(\frac{\sqrt{15}}{4}t\right) \] Because of the factor \(e^{-5s}\), the time-shift property gives \(y(x) = g(x-5)\) for \(x>5\) (and \(y(x)=0\) for \(x<5\)).

Step 4: Evaluate at \(x=10\) and check the wrong options.
\[ y(10) = g(5) = \frac{2}{\sqrt{15}}e^{-5/4}\sin\left(\frac{5\sqrt{15}}{4}\right) \] Options (A) and (C) use \(\dfrac{10}{\sqrt{15}}\) with argument \(\dfrac{\sqrt{15}}{4}\), mixing up the coefficient and forgetting to shift the argument by the delay of 5, so they are wrong. Options (C) and (D) use a growing exponential \(e^{5/4}\) instead of the decaying \(e^{-5/4}\) that this damped system must have, so they are also wrong.

Final Answer:
\[ \boxed{y(10)=\frac{2}{\sqrt{15}}e^{-5/4}\sin\left(\frac{5\sqrt{15}}{4}\right)} \]