Q.Show that the differential equation $2y\,e^{x/y}\,dx + \left(y - 2x\,e^{x/y}\right)dy = 0$ is homogeneous and find its particular solution, given that, $x = 0$ when $y = 1$.

Concept: Homogeneous Differential Equation — an equation of the form $M\,dx + N\,dy = 0$ where $M$ and $N$ are homogeneous functions of the same degree.

Step 1: Check homogeneity

Here $M = 2y e^{x/y}$, $N = y - 2x e^{x/y}$. Replace $x$ by $tx$, $y$ by $ty$:

$M(tx, ty) = 2(ty) e^{tx/(ty)} = t \cdot 2y e^{x/y} = t M(x,y)$

$N(tx, ty) = ty - 2(tx) e^{tx/(ty)} = t(y - 2x e^{x/y}) = t N(x,y)$

Both are degree 1 → homogeneous.

Step 2: Substitute $x = vy$, so $dx = v\,dy + y\,dv$

The equation becomes:

$2y e^{v} (v\,dy + y\,dv) + (y - 2vy e^{v})\,dy = 0$

Simplify: $(2vy e^{v} + y - 2vy e^{v})\,dy + 2y^2 e^{v}\,dv = 0$

$\Rightarrow y\,dy + 2y^2 e^{v}\,dv = 0$

Divide by $y$ ($y \neq 0$): $dy + 2y e^{v}\,dv = 0$

Step 3: Separate and integrate …

Homogeneity. Treating $x$ as the dependent variable,

which depends only on $x/y$, so the equation is homogeneous.

Substitute $x=vy$, giving $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$ and $x/y=v$:

Separate and integrate.

$e^{v}\,dv=-\frac{1}{2y}\,dy \;\Longrightarrow\; \int e^{v}\,dv=-\frac12\int\frac{dy}{y} \;\Longrightarrow\; e^{v}=-\frac12\log|y|+C.$ …

Method: Homogeneous equation solved with $x = vy$

Use this when the equation is homogeneous but the natural grouping is $\dfrac{x}{y}$ rather than $\dfrac{y}{x}$ — a tell is a term like $e^{x/y}$. Substituting $x = vy$ (instead of $y = vx$) keeps the algebra clean.

Steps

Step 1: Confirm homogeneity and choose the convenient variable.

Check that $M$ and $N$ share a degree. When $\dfrac{x}{y}$ is what appears, treat $x$ as the dependent variable and set $x = vy$, so $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$ and $\dfrac{x}{y} = v$.

Step 2: Substitute and cancel. …

Common Mistakes

Mistake 1: Substituting $y = vx$ when $x = vy$ is the natural choice.

Why it's wrong: the term $e^{x/y}$ makes $\dfrac{x}{y}$ the sensible ratio; forcing $y = vx$ produces awkward $e^{1/v}$ expressions. Correct approach: set $x = vy$ so $\dfrac{x}{y} = v$ and $e^{x/y} = e^{v}$.

Mistake 2: Mishandling $dx$ under the substitution.

Why it's wrong: with $x = vy$, $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$; dropping the $y\dfrac{dv}{dy}$ term breaks the separation. Correct approach: differentiate the product correctly. …