Animated Solution for Mathematics - Differential Equations: If $y=y(x)$ is the solution of the differential equation $\frac{5+e^x}{2+y}\frac{dy}{dx}+e^x=0$ satisfying $y(0)=1$ then a value of $y({\log }_{e}13)$ is :

Welcome, fellow traveler, to the elegant world of differential equations. Today, we are not just solving a math problem; we are uncovering the hidden relationship between two variables, $x$ and $y$, governed by the equation:

Imagine this equation as a complex dance where $x$ and $y$ are partners. Our goal is to find the path they trace together, starting from the point $(0,1)$.

In the realm of differential equations, the first step is often the most critical. We want to group our partners by moving all $y$ terms to one side and all $x$ terms to the other.

First, let's shift the $e^x$ term to the right side:

With a graceful rearrangement, we separate the variables:

Look at how beautifully they have parted ways! We have successfully isolated the variables, setting the stage for the next act.

Now that our variables are separated, we invite the integral sign to the party:

On the left, the integral of $\frac{1}{2+y}$ is a standard result: $\ln |2+y|$. On the right, we encounter a classic pattern.

Notice that the derivative of the denominator, $5+e^x$, is exactly the numerator, $e^x$. This is the hallmark of the logarithmic integral $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|$.

Thus, the right side becomes $-\ln |5+e^x|+\ln C$. We choose $\ln C$ as our constant of integration to make our calculations cleaner.

We now have the equation $\ln |2+y|=-\ln |5+e^x|+\ln C$. Let's bring the logarithmic terms together:

Using the logarithmic property $\ln a+\ln b=\ln (ab)$, we combine them into:

By exponentiating both sides, we arrive at the compact, elegant general solution:

This equation represents the entire family of curves that satisfy our differential equation.

We are not looking for just any curve; we are looking for the one that passes through $(0,1)$. By substituting $x=0$ and $y=1$ into our general solution, we get:

Since $e^0=1$, this simplifies to $3\cdot 6=C$, which means $C=18$. Our particular solution is now locked in:

Finally, we seek the value of $y$ when $x=\ln 13$. Substituting this into our particular solution, we get:

Using the identity $e^{\ln a}=a$, we simplify $e^{\ln 13}$ to $13$. The equation becomes:

Dividing both sides by $18$, we find $2+y=1$. This leads us to the final answer: