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: Write the characteristic equation.
The equation \(y''+y'+y=0\) has constant coefficients, so we look for solutions of the form \(y=e^{rx}\). Substituting gives the characteristic equation
\[ r^2+r+1=0 \]

Step 2: Solve for the roots.
Using the quadratic formula,
\[ r=\frac{-1\pm\sqrt{1-4}}{2}=\frac{-1\pm\sqrt{-3}}{2}=-\frac{1}{2}\pm i\frac{\sqrt3}{2} \]
The roots are complex, with real part \(-\dfrac12\) and imaginary part \(\pm\dfrac{\sqrt3}{2}\).

Step 3: Write the general solution.
For complex roots \(\alpha\pm i\beta\), the general solution has the form
\[ y(x)=e^{\alpha x}\left(C_1\cos(\beta x)+C_2\sin(\beta x)\right) \]
Here \(\alpha=-\dfrac12\) and \(\beta=\dfrac{\sqrt3}{2}\), so
\[ y(x)=e^{-x/2}\left(C_1\cos\left(\frac{\sqrt3\,x}{2}\right)+C_2\sin\left(\frac{\sqrt3\,x}{2}\right)\right) \]

Step 4: Apply the first initial condition.
At \(x=0\), \(\cos(0)=1\) and \(\sin(0)=0\), so
\[ y(0)=C_1=1 \]
Thus \(C_1=1\).

Step 5: Differentiate to apply the second initial condition.
\[ y'(x)=-\frac12e^{-x/2}\left(C_1\cos\left(\frac{\sqrt3\,x}{2}\right)+C_2\sin\left(\frac{\sqrt3\,x}{2}\right)\right)+e^{-x/2}\left(-C_1\frac{\sqrt3}{2}\sin\left(\frac{\sqrt3\,x}{2}\right)+C_2\frac{\sqrt3}{2}\cos\left(\frac{\sqrt3\,x}{2}\right)\right) \]
At \(x=0\), this simplifies to
\[ y'(0)=-\frac12C_1+\frac{\sqrt3}{2}C_2 \]

Step 6: Solve for \(C_2\).
We are given \(y'(0)=1\) and we know \(C_1=1\), so
\[ 1=-\frac12(1)+\frac{\sqrt3}{2}C_2 \]
\[ \frac32=\frac{\sqrt3}{2}C_2 \]
\[ C_2=\frac{3}{\sqrt3}=\sqrt3 \]

Step 7: Write the final solution.
With \(C_1=1\) and \(C_2=\sqrt3\),
\[ \boxed{y(x)=e^{-x/2}\left(\cos\left(\frac{\sqrt3\,x}{2}\right)+\sqrt3\sin\left(\frac{\sqrt3\,x}{2}\right)\right)} \]
Hence, the correct option is (A).