Let the solution curve of the differential equation x dy/dx - y = √(y² + 16x²), y(1) = 3 be y = y(x)

Q: Why do we call this a homogeneous equation? It doesn't look like $f(x,y) = f(kx, ky)$.

Actually, it does! If you replace $x$ with $kx$ and $y$ with $ky$, the $k$ factors cancel out. That's the secret definition of homogeneity. The degrees of all terms are consistent, allowing the $v$ substitution to work.

Q: Why do we have to isolate the radical before squaring? Can't I just square both sides immediately?

If you square immediately, you get cross-terms like $2y\sqrt{y^2+64}$, which keeps the radical alive. Isolating it first is the strategic move to eliminate the root entirely and simplify the equation into a linear one.

Q: Is there a faster way than the $y=vx$ substitution?

While you could try other methods, $y=vx$ is the most robust path for this structure. It turns a complex problem into a standard integral. Trust the process!

Question

Let the solution curve of the differential equation $x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}, y(1) = 3$ be $y = y(x)$. Then $y(2)$ is equal to :

Options:

(A) 15

(B) 11

(C) 13

(D) 17

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a differential equation that, at first glance, might seem like a daunting fortress of square roots and variables.

Let us look at our equation:

$$x \frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$$

When you see a structure like this, do not panic. Instead, look at the degrees. The term $x$ is degree one, $y$ is degree one, and inside the radical, $y^2$ and $x^2$ are both degree two.

When you take the square root of a degree-two term, you get degree one. Everything is balanced. This is the hallmark of a **homogeneous differential equation**, a beautiful symmetry that we can exploit.

### The Dance of Substitution

To unlock this, we use our most trusted key: the substitution $y = vx$. This is not just a random choice; it is a transformation that shifts our perspective.

If $y = vx$, then by the product rule, the derivative $\frac{dy}{dx}$ becomes:

$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$

Now, we substitute these into our original equation. The left side becomes $x(v + x \frac{dv}{dx}) - vx$, and the right side becomes $\sqrt{(vx)^2 + 16x^2}$.

Expanding the left side gives us $xv + x^2 \frac{dv}{dx} - vx$. The $xv$ and $-vx$ terms cancel out perfectly, leaving us with:

$$x^2 \frac{dv}{dx} = x\sqrt{v^2 + 16}$$

### The Integration Journey

We have simplified the equation significantly. Dividing both sides by $x$, we get:

$$x \frac{dv}{dx} = \sqrt{v^2 + 16}$$

Now, the variables are separated. We can write this as:

$$\frac{dv}{\sqrt{v^2 + 16}} = \frac{dx}{x}$$

This is where the beauty of calculus shines. We integrate both sides. The right side is a simple $\ln |x| + \ln C$.

For the left side, we use the standard integral formula $\int \frac{du}{\sqrt{u^2 + a^2}} = \ln |u + \sqrt{u^2 + a^2}|$. Applying this, we get:

$$\ln |v + \sqrt{v^2 + 16}| = \ln |Cx|$$

### The Final Reveal

By exponentiating both sides, we remove the logarithms:

$$v + \sqrt{v^2 + 16} = Cx$$

Now, we must return to our original variables. Substituting $v = \frac{y}{x}$ back in, we get $\frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 16} = Cx$. Multiplying by $x$ yields the general solution:

$$y + \sqrt{y^2 + 16x^2} = Cx^2$$

We are given the initial condition $y(1) = 3$. Plugging these values in, we find $3 + \sqrt{9 + 16} = C(1)^2$, which simplifies to $3 + 5 = C$, so $C = 8$.

Our specific curve is $y + \sqrt{y^2 + 16x^2} = 8x^2$. To find $y(2)$, we set $x = 2$:

$$y + \sqrt{y^2 + 64} = 8(4) = 32$$

Isolating the radical gives $\sqrt{y^2 + 64} = 32 - y$. Squaring both sides, we get $y^2 + 64 = 1024 - 64y + y^2$.

The $y^2$ terms vanish, leaving $64y = 960$. Thus, the final answer is **$y = 15$**.

Animated Solution for Mathematics - Differential Equations: Let the solution curve of the differential equation $x \frac{d y}{d x} - y = y^{2} + 16 x^{2}, y (1) = 3$ be $y = y (x)$ . Then $y (2)$ is equal to :

Select Answer:

Visualized Solution

Identify the Equation Type

The Homogeneous Substitution

Substitute into the DE

Simplify the Expression

Variable Separable Form

Integrate Both Sides

Combine Logarithms & Back-Substitute

Simplify the General Solution

Apply Initial Condition

Find the Constant C

Substitute x = 2

Isolate the Radical and Square

Final Calculation

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Dance of Substitution

The Integration Journey

The Final Reveal

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