Question

Consider the second-order differential equation
\[ \frac{d^2y}{dx^2}+\frac{dy}{dx}+y=0 \]
with initial conditions
\[ y(0)=1,\quad \left.\frac{dy}{dx}\right|_{x=0}=1 \]
The solution is given by

• A. \(y(x)=\exp(-x/2)\left(\cos\left(\dfrac{\sqrt3\,x}{2}\right)+\sqrt3\sin\left(\dfrac{\sqrt3\,x}{2}\right)\right)\)
• B. \(y(x)=\exp(-x/2)\left(\cos\left(\dfrac{\sqrt3\,x}{2}\right)+\dfrac{1}{\sqrt3}\sin\left(\dfrac{\sqrt3\,x}{2}\right)\right)\)
• C. \(y(x)=\exp(-x/2)\left(\cos\left(\dfrac{\sqrt3\,x}{2}\right)-\dfrac{1}{\sqrt3}\sin\left(\dfrac{\sqrt3\,x}{2}\right)\right)\)
• D. \(y(x)=\exp(-x/2)\left(\cos\left(\dfrac{\sqrt3\,x}{2}\right)-\sqrt3\sin\left(\dfrac{\sqrt3\,x}{2}\right)\right)\)

Hint:

Find the roots of the characteristic equation first, then fix the two constants using the two given initial conditions.

Answer 1

The Correct Option is A

Step 1: Take the Laplace transform of the equation.
Apply the Laplace transform to $y''+y'+y=0$ using $\mathcal{L}\{y''\}=s^2Y(s)-sy(0)-y'(0)$ and $\mathcal{L}\{y'\}=sY(s)-y(0)$. With $y(0)=1$ and $y'(0)=1$,
\[ s^2Y(s)-s-1+sY(s)-1+Y(s)=0 \]

Step 2: Collect terms in $Y(s)$.
\[ Y(s)(s^2+s+1)=s+2 \]
\[ Y(s)=\frac{s+2}{s^2+s+1} \]

Step 3: Complete the square in the denominator.
\[ s^2+s+1=\left(s+\frac12\right)^2+\frac34 \]

Step 4: Rewrite the numerator around the same shift.
Let $p=s+\dfrac12$, so $s=p-\dfrac12$ and $s+2=p+\dfrac32$. Then
\[ Y(s)=\frac{p+\frac32}{p^2+\frac34}=\frac{p}{p^2+\frac34}+\frac{3/2}{p^2+\frac34} \]

Step 5: Match to standard Laplace pairs.
With $\omega=\dfrac{\sqrt3}{2}$, $\mathcal{L}^{-1}\left\{\dfrac{p}{p^2+\omega^2}\right\}=\cos(\omega x)$ and $\mathcal{L}^{-1}\left\{\dfrac{\omega}{p^2+\omega^2}\right\}=\sin(\omega x)$, both shifted by $e^{-x/2}$ because of the $p=s+\frac12$ substitution. The second term becomes $\dfrac{3/2}{p^2+\omega^2}=\dfrac{3/2}{\omega}\cdot\dfrac{\omega}{p^2+\omega^2}$, and $\dfrac{3/2}{\omega}=\dfrac{3/2}{\sqrt3/2}=\sqrt3$.

Step 6: Write the inverse transform.
\[ y(x)=e^{-x/2}\left(\cos\left(\frac{\sqrt3}{2}x\right)+\sqrt3\sin\left(\frac{\sqrt3}{2}x\right)\right) \]
This matches option (A) exactly.
\[ \boxed{y(x)=e^{-x/2}\left(\cos\left(\frac{\sqrt3\,x}{2}\right)+\sqrt3\sin\left(\frac{\sqrt3\,x}{2}\right)\right)} \]