Q.Show that the family of curves for which the slope of the tangent at any point $(x, y)$ on it is $\frac{x^2 + y^2}{2xy}$, is given by $x^2 - y^2 = cx$.

The key idea is that the slope condition gives a homogeneous differential equation — the right-hand side is a function of $y/x$ alone.

Step 1 – Write the differential equation

The slope is $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$. Divide numerator and denominator by $x^2$:

Step 2 – Substitute $y = vx$

Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$. The equation becomes:

Step 3 – Separate variables

Subtract $v$ from both sides:

Separate:

Step 4 – Integrate

Integrate both sides: …

The problem gives us the slope of the tangent at any point $(x, y)$ on a curve:

We need to show that the family of curves satisfying this is $x^2 - y^2 = cx$.

Why the homogeneous approach works

Look at the right-hand side: both numerator and denominator are homogeneous of degree 2 — each term is $x^2$, $y^2$, or $xy$. That means the whole fraction can be written as a function of $y/x$ alone. When a differential equation has this property, the substitution $y = vx$ (where $v = y/x$) always works, because it turns the equation into one where variables separate cleanly.

1. Rewrite the equation in terms of $v = y/x$

   Let $y = vx$, so $v$ is a new function of $x$. Then:

Substitute into the given equation:

2. Separate the variables

   Subtract $v$ from both sides:

So:

Now separate:

3. Integrate both sides

   The left side integrates nicely with a substitution. Let $u = 1 - v^2$, then $du = -2v\,dv$, so $2v\,dv = -du$. Hence:

The right side is:

Combining constants:

where $C = C_2 - C_1$.

4. Solve for $v$ …

Method: Homogeneous slope condition leading to a family of curves

Use this when a slope $\dfrac{dy}{dx}$ is given as a ratio of same-degree expressions in $x$ and $y$ and you must derive the family of curves that satisfies it.

Steps

Step 1: Write the slope as $\dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right)$.

Divide top and bottom by the common power of $x$ so only $\dfrac{y}{x}$ remains — confirming the equation is homogeneous.

Step 2: Substitute $y = vx$, $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$, and subtract $v$.

This gives $x\dfrac{dv}{dx} = (\text{function of }v)$.

Step 3: Separate and integrate, watching for a log-substitution. …

Common Mistakes

Mistake 1: Forgetting to subtract $v$ after substituting $y = vx$.

Why it's wrong: you must form $x\dfrac{dv}{dx} = \dfrac{1+v^2}{2v} - v = \dfrac{1-v^2}{2v}$; omitting the $-v$ gives the wrong separable equation. Correct approach: always move the stray $v$ across.

Mistake 2: Sign error integrating $\dfrac{2v}{1-v^2}$.

Why it's wrong: with $u = 1-v^2$, $du = -2v\,dv$, so $\displaystyle\int\frac{2v}{1-v^2}\,dv = -\log|1-v^2|$ — the minus sign is easy to lose. Correct approach: track the substitution's sign carefully. …