Q.$\left(1+e^{\frac{x}{y}}\right) dx + e^{\frac{x}{y}} \left(1-\frac{x}{y}\right) dy = 0$

The key idea is that this is a homogeneous differential equation — it can be rewritten in terms of $\frac{x}{y}$.

Step 1: Rewrite in standard form

The right-hand side depends only on $\frac{x}{y}$, confirming homogeneity.

Step 2: Substitute $x = v y$, so $\frac{dx}{dy} = v + y\frac{dv}{dy}$.

The equation becomes:

Step 3: Separate variables

Thus:

Step 4: Integrate both sides

$\int \frac{1 + e^{v}}{v + e^{v}} \, dv = -\log|y| + C$ …

The given equation is:

We can rewrite it as:

The right-hand side is a function of $\frac{x}{y}$ alone. This is the hallmark of a homogeneous differential equation (in the sense that the function $f(x,y)$ is homogeneous of degree zero). When you see an equation where every term has the same total degree in $x$ and $y$, or where the ratio $x/y$ appears naturally, the substitution $x = vy$ (or $y = ux$) is the standard path.

Why does this work? Because if $x = vy$, then $\frac{dx}{dy} = v + y\frac{dv}{dy}$. The original equation, which is messy in $x$ and $y$, becomes a separable equation in $v$ and $y$. That is the entire point: turn a complicated coupled pair into something you can integrate.

Let’s do it step by step.

1. Substitute $x = vy$.

   Then $\frac{dx}{dy} = v + y\frac{dv}{dy}$. Also, $\frac{x}{y} = v$.

   The equation becomes:

2. Isolate the derivative term. Bring $v$ to the right:

Combine the right-hand side into a single fraction:

Simplify the numerator:

So we have:

3. Separate variables. Multiply both sides by $dy$ and divide by the $v$-expression:

4. Integrate both sides. …

Method: Homogeneous equation with $e^{x/y}$ — use $x=vy$

Use this when the exponential is $e^{x/y}$ (as in $(1+e^{x/y})\,dx+e^{x/y}(1-\frac{x}{y})\,dy=0$); since $\frac{x}{y}$ is the natural ratio, substitute $x=vy$.

Steps

Step 1: Substitute $x=vy$

With $v=\dfrac{x}{y}$ and $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$, every $e^{x/y}$ becomes $e^{v}$.

Step 2: Separate and spot the $\frac{g'}{g}$ pattern

After simplifying, the $v$-side takes the form $\dfrac{d(v+e^{v})}{v+e^{v}}$, whose integral is a logarithm. …

Common Mistakes

Mistake 1: Substituting $y=vx$ when the exponent is $x/y$

Why it's wrong: $e^{x/y}$ becomes clean only under $x=vy$ (giving $e^{v}$); $y=vx$ makes it $e^{1/v}$, which is far harder. Correct approach: match the substitution to the ratio in the exponent.

Mistake 2: Missing the $\dfrac{d(v+e^v)}{v+e^v}$ pattern …