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

A different way to solve this is to use the jump condition that an impulse creates in the derivative, instead of using the Laplace transform.

For $x<5$, there is no forcing yet and the initial conditions are $y(0)=0$, $y'(0)=0$, so the unique solution of the homogeneous equation on that interval is simply $y(x)=0$.

Integrate $2y''+y'+2y=\delta(x-5)$ across a tiny interval around $x=5$, from $5-\epsilon$ to $5+\epsilon$, and let $\epsilon\to 0$. The terms $\int y'\,dx$ and $\int 2y\,dx$ shrink to 0 because $y$ and $y'$ stay bounded, but $\int 2y''\,dx = 2[y'(5^+)-y'(5^-)]$ must equal $\int \delta(x-5)\,dx=1$.

Since $y'(5^-)=0$ (from the zero solution before the impulse), this gives $y'(5^+) = \frac{1}{2}$. Also $y$ stays continuous at $x=5$, so $y(5^+)=y(5^-)=0$.

For $x>5$, let $\tau=x-5$. The equation becomes homogeneous again, $2y''+y'+2y=0$, with fresh initial data $y(0)=0$ and $y'(0)=\frac12$ in terms of $\tau$.

The characteristic equation is $2r^2+r+2=0$, giving $r = \dfrac{-1\pm\sqrt{1-16}}{4} = -\dfrac{1}{4} \pm i\dfrac{\sqrt{15}}{4}$.

So the general solution is $y(\tau) = e^{-\tau/4}\left[C_1\cos\left(\dfrac{\sqrt{15}}{4}\tau\right)+C_2\sin\left(\dfrac{\sqrt{15}}{4}\tau\right)\right]$.

Using $y(0)=0$ gives $C_1=0$. Differentiating and using $y'(0)=\frac12$ gives $C_2\cdot\dfrac{\sqrt{15}}{4}=\dfrac12$, so $C_2=\dfrac{2}{\sqrt{15}}$.

So $y(\tau) = \dfrac{2}{\sqrt{15}}e^{-\tau/4}\sin\left(\dfrac{\sqrt{15}}{4}\tau\right)$, and at $x=10$ we have $\tau=5$, giving

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

Let's summarize:

• Before the impulse hits at $x=5$, the response is zero because the system starts at rest.
• The impulse instantly changes the derivative $y'$ by $\frac{1}{2}$ at $x=5$, and after that the system just follows its natural damped oscillation.

This matches the Laplace transform result exactly, confirming option (B).

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