Q.Find the equation of a curve passing through $(2,\,1)$ if the slope of the tangent to the curve at any point $(x,\,y)$ is $\frac{x^2+y^2}{2xy}$.

The slope equation $\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$ is homogeneous (numerator and denominator are both degree $2$). Put $y=vx$, so $\frac{dy}{dx}=v+x\frac{dv}{dx}$:

Separate and integrate, noting $\int\frac{2v}{1-v^2}\,dv=-\log|1-v^2|$: …

Recognise the type

The tangent slope is $\frac{dy}{dx}=\frac{x^2+y^2}{2xy}$. Dividing top and bottom by $x^2$ makes it a function of $y/x$ alone:

That marks it as a homogeneous equation, solved by $y=vx$.

Substitute

With $y=vx$, $\frac{dy}{dx}=v+x\frac{dv}{dx}$ and $y/x=v$:

Subtract $v$:

Separate and integrate

Since the numerator is $-\frac{d}{dv}(1-v^2)$, the left integral is $-\log|1-v^2|$: …

Method: Homogeneous slope-DE for a curve through a point

Use this when the slope $\dfrac{dy}{dx}$ is a ratio of same-degree expressions in $x,y$ and a point is given.

Steps

Step 1: Confirm homogeneity and substitute $y=vx$

A ratio like $\dfrac{x^2+y^2}{2xy}$ is degree-$0$ homogeneous. Put $y=vx$, $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.

Step 2: Separate in $v$ and $x$ …

Common Mistakes

Mistake 1: Not recognising the slope as homogeneous

Why it's wrong: $\frac{x^2+y^2}{2xy}$ is degree-$0$ homogeneous, so $y=vx$ applies. Correct approach: substitute $y=vx$.

Mistake 2: Mis-integrating $\frac{2v}{1-v^2}$

Why it's wrong: it equals $-\log|1-v^2|$ (numerator is $-$ derivative of denominator). Correct approach: use the $\frac{f'}{f}$ pattern with the sign. …