If y dy/dx = x [y²/x² + φ(y²/x²)/φ'(y²/x²)], x 0, φ 0, and y(1) = -1, then φ(y²/4) is equal to :

Q: Wait, why do we call this a homogeneous equation? It doesn't look like the standard $dy/dx = f(y/x)$ form immediately.

That is a sharp observation! While it isn't explicitly written as $dy/dx = F(y/x)$, the structure $y \frac{dy}{dx} = x [ ... ]$ can be rearranged by dividing by $x$ to get $\frac{dy}{dx} = \frac{x}{y} [ ... ]$. More importantly, the presence of the ratio $y^2/x^2$ inside the $\phi$ functions is the 'signature' of a homogeneous equation. In JEE Advanced, you must learn to recognize these patterns even when they are slightly disguised.

Q: Why did we choose $u = \phi(t^2)$ for the integral? Is there a trick to seeing that?

It is not a trick, but a fundamental pattern recognition skill. Whenever you see a function and its derivative in the same integral, like $\phi'(t^2)$ and $\phi(t^2)$, your brain should immediately flag it for $u$-substitution. Since the derivative of $\phi(t^2)$ involves $2t \phi'(t^2)$ (by the chain rule), and we had a $t \phi'(t^2)$ term in our numerator, the substitution was the natural path to simplification.

Q: I got $\phi(t^2) = kx^2$, but how do I know what to do with the initial condition $y(1) = -1$?

The initial condition is your key to unlocking the constant $k$. The condition $y(1) = -1$ tells us that when $x=1$, $y=-1$. Since $t = y/x$, at this specific point, $t = -1/1 = -1$, so $t^2 = 1$. By plugging these values into your general solution, you isolate $k$. Always remember: the general solution describes a family of curves; the initial condition picks the specific one that passes through your point.

Question

If $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(\frac{y^2}{x^2})}{\phi'(\frac{y^2}{x^2})} \right], x > 0, \phi > 0$, and $y(1) = -1$, then $\phi(\frac{y^2}{4})$ is equal to :

Options:

(A) $4\phi(2)$

(B) $4\phi(1)$

(C) $2\phi(1)$

(D) $\phi(1)$

Sigma Percentile · Accepted Answer

### The Art of Seeing Through Complexity

Welcome, fellow traveler on the JEE Advanced journey. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of notation. You see the symbols $\phi$ and $\phi'$, and your instinct might be to panic.

But I want you to take a deep breath. In mathematics, especially in differential equations, complexity is often just a mask. Our job is to peel back that mask and reveal the elegant, simple structure underneath.

### Phase 1: The Homogeneous Insight

Let us look at the given equation:

$$y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(\frac{y^2}{x^2})}{\phi'(\frac{y^2}{x^2})} \right]$$

Do you see it? The term $\frac{y^2}{x^2}$ appears repeatedly. This is not a coincidence; it is a signal.

In the world of differential equations, when you see the ratio of variables appearing consistently, you are almost certainly dealing with a **Homogeneous Differential Equation**. This is your first victory. By recognizing this, you have already decided on your strategy: we are going to transform it into a language we understand.

### Phase 2: The Substitution Dance

To simplify this, we introduce the classic substitution: $y = tx$. This is the key that unlocks the door.

But remember, we cannot just change $y$ and leave $\frac{dy}{dx}$ alone. We must respect the rules of calculus. Differentiating $y = tx$ with respect to $x$ using the product rule gives us:

$$\frac{dy}{dx} = t + x \frac{dt}{dx}$$

Now, let us substitute these into our original equation. The left side becomes $tx(t + x \frac{dt}{dx})$, and the right side becomes $x [ t^2 + \frac{\phi(t^2)}{\phi'(t^2)} ]$.

Since the problem guarantees $x > 0$, we can divide by $x$ without fear. Expanding the left side, we get:

$$t^2 + xt \frac{dt}{dx} = t^2 + \frac{\phi(t^2)}{\phi'(t^2)}$$

Look at that! The $t^2$ terms on both sides vanish like magic. We are left with $xt \frac{dt}{dx} = \frac{\phi(t^2)}{\phi'(t^2)}$. This is the moment where the problem shifts from 'terrifying' to 'solvable'.

### Phase 3: The Integration Symphony

Now, we separate the variables. We want all the $t$ terms on one side and all the $x$ terms on the other. Rearranging, we get:

$$\frac{t \phi'(t^2)}{\phi(t^2)} dt = \frac{dx}{x}$$

This is where many students stumble, but you won't. Look at the left side; we have $\phi(t^2)$ in the denominator and $\phi'(t^2)$ in the numerator. This is a classic setup for $u$-substitution.

Let $u = \phi(t^2)$. Then, by the chain rule, $du = \phi'(t^2) \cdot 2t \cdot dt$. This means $t \phi'(t^2) dt = \frac{1}{2} du$.

Substituting this back, the integral becomes:

$$\frac{1}{2} \int \frac{du}{u} = \int \frac{dx}{x}$$

Integrating both sides gives us $\frac{1}{2} \ln \phi(t^2) = \ln x + \ln c$. Using the properties of logarithms, we can simplify this to $\ln \sqrt{\phi(t^2)} = \ln (cx)$.

Removing the logarithms, we arrive at $\sqrt{\phi(t^2)} = cx$, or simply $\phi(t^2) = kx^2$, where $k = c^2$.

### Phase 4: The Final Reveal

We are almost there. We have the general solution $\phi(\frac{y^2}{x^2}) = kx^2$. Now, we apply the initial condition $y(1) = -1$.

At $x=1$, $y=-1$, so $t^2 = (\frac{-1}{1})^2 = 1$. Plugging these into our equation, we get $\phi(1) = k(1)^2$, which means $k = \phi(1)$.

Our specific solution is $\phi(\frac{y^2}{x^2}) = \phi(1)x^2$. The question asks for $\phi(\frac{y^2}{4})$. By comparing the arguments, we see that $x^2$ must be $4$.

Substituting $x^2 = 4$ into our equation, we get $\phi(\frac{y^2}{4}) = \phi(1) \cdot 4$.

And there it is: **$4\phi(1)$**. You didn't just solve a problem; you navigated a complex landscape of calculus, substitution, and logic. Remember this feeling—the feeling of turning chaos into order. That is the true essence of mathematics.

Animated Solution for Mathematics - Differential Equations: If $y \frac{d y}{d x} = x [\frac{y ^{2}}{x ^{2}} + \frac{ϕ ( \frac{y ^{2}}{x ^{2}} )}{ϕ ^{'} ( \frac{y ^{2}}{x ^{2}} )}], x > 0, ϕ > 0$ , and $y (1) = - 1$ , then $ϕ (\frac{y ^{2}}{4})$ is equal to :

Select Answer:

Visualized Solution

Identify the Homogeneous Form

Define the Substitution $y = t x$

Substitute into the Original Equation

Simplify the Equation

Separate the Variables

Integrate Both Sides

Solve the Integrals

Find the General Solution

Apply the Initial Condition

Final Evaluation

The Sigma Insight: Homogeneous Differential Equations

The Art of Seeing Through Complexity

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ydxdy​=x[x2y2​+ϕ′(x2y2​)ϕ(x2y2​)​],x>0,ϕ>0, and y(1)=−1, then ϕ(4y2​) is equal to :