Q.Show that the differential equation $x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y\cos\left(\frac{y}{x}\right) + x$ is homogeneous and solve it.

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:

where $C = C_2 - C_1$.

5. Solve for $v$ and back-substitute Exponentiate both sides:

Let $K = \pm e^C$ (an arbitrary constant), so:

Recall $v = y/x$, so:

Renaming the constant $K$ as $C$ (since it’s arbitrary), we get:

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 $y$ Since $v = y/x$, we have $v^2 = y^2/x^2$, so:

Thus:

7. Exponentiate to get the final form

Let $c = \pm e^C$ (absorbing the absolute value), we obtain:

This matches option (A).

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:

The $-\log|x|$ on the left and the $\log|x|$ on the right cancel:

6. Rewrite in exponential form Multiply by 2:

Let $2C = \log c$ (absorbing constant):

Exponentiate:

Take square root (since $c$ is arbitrary, we can rename $\sqrt{c}$ as a new constant $c$):

[!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:

Multiplying by $2$ gives $\log(x^2+y^2)=2\tan^{-1}\tfrac{y}{x}+c$.

[!ANSWER]

(C) $\log(x^2+y^2)=2\tan^{-1}\dfrac{y}{x}+c$

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)

=> [phi'(v)/phi(v)] dv = 2 dx/x.

Integrate:

log|phi(v)| = 2 log|x| + log C

=> phi(v) = C x^2

=> phi(y^2/x^2) = C x^2.

ANSWER: C

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:

This is homogeneous of degree 2.

4. Option (D): $f(x,y) = 2x - y$ Replace:

This is homogeneous of degree 1.

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 $C = -\frac12$:

Multiply both sides by 2:

This matches option (C) exactly.