This problem aims to familiarize us with **higher-order differential** equations. The concept required to solve this problem is **ordinary differential equations** given at a specific point and **product rule.** Here we are going to find the **second-order** differential with the help of a **reference** point.

Now, an **ordinary differential** **equation** also known as **ODE** is an equation that implicates ordinary **derivatives** which are the opposite of **partial derivatives** of a function. Usually, our purpose is to minimize an **ODE,** to resolve what function or functions fulfill the **equation.**

For this particular problem, we are dealing with **second order differential** equation which is of the form $y“ + p(x)y` + q(x)y = f(x)$. This equation contains some **constant coefficients** only if the functions $p(x)$ and $q(x)$ are constants.

## Expert Answer

We are given an **equation:**

\[ xy + 3e^y = 3e \space (Eq.1) \]

Where $e$ is a **constant** value.

At $x = 0$, $y$ comes out to be:

\[ (0)y + 3e^y = 3e \]

\[ 3e^y = 3e \]

\[ e^y = e \]

\[ y = 1 \]

Now, **d****ifferentiating** both sides of the equation $Eq.1$ with respect to $x$:

\[ \dfrac{d(xy + 3e^y)}{dx} = \dfrac{d(3e)}{dx} \]

\[ \dfrac{d(xy)}{dx} + \dfrac{d(3e^y)}{dx} = \dfrac{d(3e)}{dx} \]

Let $\dfrac{d(xy)}{dx} = I$, solving this **equation** using the **product rule** which is basically of the form:

\[ f(x) = u(x)\times v(x) \]

Then,

\[ f'(x) = u'(x).v(x) + u(x).v'(x) \]

**Solving** $I$:

\[ I = \dfrac{d(xy)}{dx} \]

\[ I = x \dfrac{dy}{dx} + y \dfrac{dx}{dx} \]

\[ I = x \dfrac{dy}{dx} + y \]

Plugging $I$ back into the **main equation** gives us:

\[ x \dfrac{dy}{dx} + 1 + 3e \dfrac{dy}{dx} = 0 \]

Taking $\dfrac{dy}{dx}$ common:

\[ \dfrac{dy(x + 3e)}{dx} = -1 \]

\[ \dfrac{dy}{dx} = \dfrac{-1}{(x + 3e)} \]

This is the **expression** for the **first-order** derivative.

At $x = 0$, $y`$ comes out to be:

\[ \dfrac{dy}{dx} = \dfrac{-1}{(0 + 3e)} \]

\[ \dfrac{dy}{dx} = \dfrac{-1}{3e} \]

Now calculating the **second-order** derivative:

\[ \dfrac{d}{dx} \times \dfrac{dy}{dx} = \dfrac{d}{dx} \times \dfrac{-1}{(x + 3e)} \]

\[ \dfrac{d^2y}{dx^2} = – \dfrac{d(x + 3e)^{-1}}{dx} \]

\[ \dfrac{d^2y}{dx^2} = \dfrac{1}{(x + 3e)^2} \]

This is our expression for the **second-order** derivative.

At $x = 0$, $y“$ comes out to be:

\[ \dfrac{d^2y}{dx^2} = \dfrac{1}{(3e)^2} \]

\[ \dfrac{d^2y}{dx^2} = \dfrac{1}{9e^2} \]

## Numerical Result

The **value** of $y“$ at **point** $x = 0$ comes out to be $ \dfrac{d^2y}{dx^2} = \dfrac{1}{9e^2} $.

## Example

If $xy + 6e^y = 6e$, find $y`$ at $x = 0$.

We are given an **equation:**

\[ xy + 6e^y = 6e \space (Eq.2)\]

At $x = 0$, $y$ comes out to be:

\[ (0)y + 6e^y = 6e\]

\[ y = 1\]

Now, **Differentiating** both sides of the **equation** $Eq.2$ with respect to $x$:

\[\dfrac{d(xy)}{dx} + \dfrac{d(6e^y)}{dx} = \dfrac{d(6e)}{dx}\]

\[ x \dfrac{dy}{dx} + 1 + 6e \dfrac{dy}{dx} = 0\]

**Rearranging:**

\[ \dfrac{dy(x + 6e)}{dx} = -1\]

\[\dfrac{dy}{dx} = \dfrac{-1}{(x + 6e)}\]

At $x = 0$, $y`$ comes out to be:

\[\dfrac{dy}{dx} = \dfrac{-1}{6e}\]