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

The correct option is:
$y\left(x\right)=exp(-x/2)(cos\left(\frac{\sqrt{3}x}{2}\right)+\sqrt{3}sin\left(\frac{\sqrt{3}x}{2}\right))$
(as shown in the first image, which corresponds to option 1).

Step-by-Step Derivation and Explanation:

Step 1: Write down the characteristic equation
The given second-order linear homogeneous differential equation is:
$$\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$$
To solve this, we assume a solution of the form y = e^rx. Substituting this into the differential equation gives the auxiliary (characteristic) equation:
$$r^2+r+1=0$$

Step 2: Find the roots of the characteristic equation
We use the quadratic formula to solve for r:
$$r=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}$$
Simplifying the term under the square root:
$$r=\frac{-1\pm \sqrt{-3}}{2}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$$
The roots are complex conjugates of the form $\alpha \pm i\beta$, where:
$\alpha =-\frac{1}{2}$ and $\beta =\frac{\sqrt{3}}{2}$.

Step 3: State the general solution
For complex roots, the general solution of the differential equation is:
$$y\left(x\right)=e^{\alpha x}(Acos(\beta x)+Bsin(\beta x\left)\right)$$
Substituting $\alpha$ and $\beta$:
$$y\left(x\right)=e^{-x/2}(Acos\left(\frac{\sqrt{3}x}{2}\right)+Bsin\left(\frac{\sqrt{3}x}{2}\right))$$

Step 4: Apply the first initial condition
We are given that $y\left(0\right)=1$. Substituting x = 0:
$$1=e^0(Acos(0)+Bsin(0\left)\right)$$
Since e⁰ = 1, cos(0) = 1, and sin(0) = 0:
$$1=1\cdot (A\cdot 1+0)\Rightarrow A=1$$

Step 5: Apply the second initial condition
We are given that the derivative at x = 0 is 1, i.e., $y^{\text{'}}\left(0\right)=1$. First, let's find the derivative of y(x) using the product rule:
$$y^{\text{'}}\left(x\right)=-\frac{1}{2}{e}^{-x/2}(cos\left(\frac{\sqrt{3}x}{2}\right)+Bsin\left(\frac{\sqrt{3}x}{2}\right))+e^{-x/2}(-\frac{\sqrt{3}}{2}sin\left(\frac{\sqrt{3}x}{2}\right)+B\frac{\sqrt{3}}{2}cos\left(\frac{\sqrt{3}x}{2}\right))$$
Evaluating this derivative at x = 0:
$$y^{\text{'}}\left(0\right)=-\frac{1}{2}\left(1\right)(1+0)+\left(1\right)(0+B\frac{\sqrt{3}}{2}\cdot 1)$$
Since $y^{\text{'}}\left(0\right)=1$:
$$1=-\frac{1}{2}+B\frac{\sqrt{3}}{2}$$
Add $\frac{1}{2}$ to both sides:
$$\frac{3}{2}=B\frac{\sqrt{3}}{2}$$
Multiplying by 2 and dividing by $\sqrt{3}$:
$$B=\frac{3}{\sqrt{3}}=\sqrt{3}$$

Step 6: Write the final solution
Substituting A = 1 and $B=\sqrt{3}$ back into our general equation:
$$y\left(x\right)=exp(-x/2)(cos\left(\frac{\sqrt{3}x}{2}\right)+\sqrt{3}sin\left(\frac{\sqrt{3}x}{2}\right))$$
This perfectly matches the mathematical expression rendered in the first image (image_0.webp).