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 problem gives a geometric condition: at any point $(x, y)$ on the curve, the slope of the tangent equals the product of the coordinates. That is,

This is a first-order differential equation. The key is to recognise it as separable — we can collect all $y$ terms on one side and all $x$ terms on the other, then integrate.

The curve must pass through $(1, 1)$, so that fixes the constant of integration. Let’s work through it.

1. Set up the differential equation The slope of the tangent at $(x, y)$ is $\frac{dy}{dx}$. The product of the coordinates is $xy$. So

2. Separate the variables Bring $y$ terms to the left and $x$ terms to the right:

(We assume $y \neq 0$; the point $(1,1)$ ensures $y>0$ here.)

3. Integrate both sides

gives

Since $y$ is positive near $(1,1)$, we can drop the absolute value:

4. Apply the initial condition The curve passes through $(1, 1)$, so $x = 1$, $y = 1$:

5. Write the particular solution Substitute $C = -\frac12$: $$$$

The key insight is that the equation is homogeneous — every term has the same degree when $y$ and $x$ are considered together. That means we can set $y = vx$ (where $v = y/x$), which turns the equation into one that separates cleanly. The presence of $\tan(y/x)$ is a dead giveaway: it’s a function of the ratio $y/x$ alone.

1. Rewrite the equation in standard form Given:

Divide through by $x$ (assuming $x \neq 0$):

This is clearly homogeneous: the right-hand side depends only on $y/x$.

2. Substitute $y = vx$ Let $v = y/x$, so $y = vx$. Then differentiate:

Substitute into the equation:

The $v$ terms cancel on both sides, leaving:

3. Separate variables Bring the $v$-terms to one side and $x$-terms to the other:

Since $\tan(v) = \frac{\sin(v)}{\cos(v)}$, we have $\frac{1}{\tan(v)} = \frac{\cos(v)}{\sin(v)}$. So:

4. Integrate both sides The left side integrates to $\log|\sin(v)|$ (because the derivative of $\sin(v)$ is $\cos(v)$):

The right side gives:

Combining constants:

$\log|\sin(v)| = \log|x| + C$ …

We start by recognizing the structure: the equation $(x-y) dy = (x+y) dx$ is homogeneous — both coefficients are homogeneous functions of degree 1. For such equations, the substitution $y = vx$ (or $x = vy$) simplifies the relationship between $x$ and $y$ into a separable form.

Why this works:

When we set $y = vx$, we get $dy = v\,dx + x\,dv$. The original equation becomes an equation in $x$ and $v$ where the variables can be separated. This is the classic method for homogeneous first-order ODEs.

1. Rewrite the equation in standard form

Divide both sides by $dx$ (assuming $dx \neq 0$):

So

2. Substitute $y = vx$ Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. The equation becomes:

3. Separate variables Subtract $v$ from both sides:

So

4. Integrate both sides Left side:

The first integral is $\tan^{-1} v$. For the second, let $u = 1+v^2$, $du = 2v\,dv$, so $\int \frac{v}{1+v^2} dv = \frac{1}{2} \log(1+v^2)$.

Thus:

5. Back-substitute $v = y/x$

Simplify the log term:

So

Plugging back:

$\tan^{-1}\left(\frac{y}{x}\right) - \left[ \frac{1}{2} \log(x^2 + y^2) - \log|x| \right] = \log|x| + C$ …

[!TLDR]

Recognise the two sides as exact differentials of $x^2+y^2$ and $\tan^{-1}(y/x)$; separating and integrating gives $\log(x^2+y^2)=2\tan^{-1}(y/x)+c$.

Concept

This CBSE Class 12 differential-equations problem is solved fastest by spotting standard exact-differential combinations: $d(x^2+y^2)=2(x\,dx+y\,dy)$ and $d\!\left(\tan^{-1}\tfrac{y}{x}\right)=\dfrac{x\,dy-y\,dx}{x^2+y^2}$.

Solution

Start from

Write each side as a differential:

Divide both sides by $x^2+y^2$:

Integrate:

$\tfrac12\log(x^2+y^2)=\tan^{-1}\tfrac{y}{x}+c_1.$ …

We are given:

Concept & Intuition

The right-hand side is a ratio where both numerator and denominator are homogeneous of degree 2 (each term has total power 2). This suggests the substitution $y = vx$ (or $v = y/x$), which turns the equation into one where variables separate. The trick is that after substitution, the $x$ and $v$ parts can be isolated, leading to an implicit relation between $y$ and $x$.

Step-by-step solution

1. Rewrite using substitution $y = vx$ Let $v = y/x$, so $y = vx$. Then

Substitute into the DE:

2. Isolate the derivative term Subtract $v$ from both sides:

3. Separate variables Multiply both sides by $dx$ and divide by the factor involving $v$:

(We assume $v \neq 0$; $v=0$ gives $y=0$, which is a trivial solution but not part of the general family.)

4. Integrate both sides Left side:

Right side:

Combine constants:

5. Simplify using logarithms Bring the log terms together:

That is:

But $vx = y$, so:

6. Replace $v$ back in terms of $x$ and $y$ …

Concept: substitute v = y^2/x^2 to reduce to a separable equation.

Given: y dy/dx = x [ y^2/x^2 + phi(v)/phi'(v) ], where v = y^2/x^2.

From v = y^2/x^2 we get y^2 = v x^2. Differentiate with respect to x:

2y (dy/dx) = x^2 (dv/dx) + 2vx

=> y (dy/dx) = (x^2/2)(dv/dx) + vx.

Substitute into the given equation:

(x^2/2)(dv/dx) + vx = xv + x * phi(v)/phi'(v)

=> (x^2/2)(dv/dx) = x * phi(v)/phi'(v) …

Concept & Intuition

A function of two variables is homogeneous of degree $n$ if scaling both inputs by the same factor $t$ multiplies the output by $t^n$. This is a powerful symmetry property: the function’s form depends only on the ratio of the variables (like $y/x$) when $n=0$, or more generally on powers of $t$ times that ratio. To test homogeneity, replace $x$ with $tx$ and $y$ with $ty$, then see if you can factor out a single power of $t$. If you cannot — because the function involves non-polynomial terms like $\sin$ or $\cos$ of the original variables (not their ratio) — it is not homogeneous.

Step-by-step reasoning

1. Option (A): $f(x,y) = \sin x - \cos y$ Replace $x$ with $tx$ and $y$ with $ty$:

There is no way to factor out a common power of $t$ because $\sin(tx) \neq t^n \sin x$ for any constant $n$ (except trivially $t=1$). The function does not scale uniformly. Hence, this is not homogeneous.

2. Option (B): $f(x,y) = \cos^2\!\left(\frac{y}{x}\right) + \frac{y}{x}$ Replace:

The $t$ cancels completely, so $f(tx, ty) = t^0 f(x,y)$. This is homogeneous of degree 0.

3. Option (C): $f(x,y) = x^2 + 2xy$ Replace: $f(tx, ty) = (tx)^2 + 2(tx)(ty) = t^2 x^2 + 2 t^2 xy = t^2 (x^2 + 2xy) = t^2 f(x,y)$ …

1. The definition being tested

The degree of a differential equation is the power of the highest-order derivative, after the equation has been made free of radicals and fractional powers in the derivatives. If it cannot be made polynomial in the derivatives, the degree is not defined. So the cube root here must be cleared first — you may not read the degree off the equation as printed.

2. The equation

3. Cube both sides

This is now a polynomial in $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$.

4. Identify order and degree

• Order = $2$ (the highest derivative present is $\dfrac{d^2y}{dx^2}$). …