Animated Solution for Mathematics - Differential Equations: Let $y(x)$ be a solution of the differential equation $(1+e^x)y^{\prime }+y{e}^x=1$. If $y(0)=2$, then which of the following statement is (are) true?

Welcome, future engineer. Today, we are going to peel back the layers of a differential equation that might look intimidating at first, but is actually a masterpiece of mathematical elegance.

The problem gives us:

At first glance, it feels like a tangled mess of exponential terms. But in the world of JEE Advanced, we don't fear the mess; we look for the order within it.

The Hidden Symmetry

Look at the left-hand side (LHS) of our equation. It is not just a random collection of terms; it is a classic setup for the product rule.

Recall that the derivative of a product $u\cdot v$ is $u^{\prime }v+u{v}^{\prime }$. If we let $u=y$ and $v=(1+e^x)$, then $u^{\prime }=y^{\prime }$ and $v^{\prime }=e^x$.

When we compute the derivative of $y(1+e^x)$, we get exactly $y^{\prime }(1+e^x)+y(e^x)$. Our entire LHS collapses into a single, neat exact derivative:

This is the 'Aha!' moment that turns a complex problem into a simple integration.

The Integration Journey

Now that we have $\frac{d}{dx}[y(1+e^x)]=1$, the path forward is clear. We integrate both sides with respect to $x$:

The integral of a derivative is simply the original function, so we get:

This is our general solution. But we are not done yet; we need to find that elusive constant $C$.

Locking the Curve

The problem provides a lifeline: the initial condition $y(0)=2$. This tells us exactly which curve we are dealing with.

By substituting $x=0$ and $y=2$ into our equation, we get:

Since $e^0=1$, this simplifies to $2(1+1)=C$, which means $C=4$. Our particular solution is now locked in:

The Calculus of Critical Points

Now, let's tackle the options. Option A asks if $y(-4)=0$.

Plugging in $x=-4$, we get:

Option A is true! For Option B, $y(-2)=\frac{-2+4}{1+e^{-2}}=\frac{2}{1+e^{-2}}$, which is clearly not zero. So, Option B is false.

Finally, we look for critical points. We need to find where $y^{\prime }(x)=0$. Using the quotient rule on $y(x)=\frac{x+4}{1+e^x}$, we find:

Simplifying the numerator, we get $1+e^x-x{e}^x-4e^x=1-(x+3)e^x$. Setting this to zero, we need to check if $f(x)=(x+3)e^x-1=0$ has a root in $(-1,0)$.

Using the Intermediate Value Theorem, we check the boundaries:

Since $e\approx 2.718$, $\frac{2}{e}<1$, so $f(-1)<0$. At the other boundary, $f(0)=(0+3)e^0-1=3-1=2$, which is greater than zero.

Since the function changes sign from negative to positive, it must cross zero somewhere in between. Thus, a critical point exists in $(-1,0)$. Option C is true!

We have successfully decoded the problem. Through symmetry, integration, and the power of the Intermediate Value Theorem, we have proven that Options A and C are the correct statements. Keep practicing, and remember: the beauty of math lies in finding the simple truth hidden behind the complex surface.