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If xy + 3ey = 3e, find the value of y” at the point where x = 0.

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, differentiating 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:

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

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