If dy/dx = xy/(x²+y²); y(1) = 1: then a value of x satisfying y(x) = e is: (A) √3e (B) e/√2 (C) √2e

Q: Wait, why do we use $y = vx$ instead of $x = vy$?

Great question! You can actually use either, but $y = vx$ is the standard convention when the equation is given as $\frac{dy}{dx} = f(x, y)$. It makes the derivative $\frac{dy}{dx}$ straightforward to compute. If you chose $x = vy$, you would be solving for $\frac{dx}{dy}$, which is perfectly valid but often leads to more algebraic friction.

Q: Why did the $\ln|x|$ terms cancel out? Did I make a mistake?

You are absolutely correct! When you substitute $v = \frac{y}{x}$ into the equation, the term $\ln|v|$ becomes $\ln|\frac{y}{x}|$, which expands to $\ln|y| - \ln|x|$. Since there is already a $-\ln|x|$ on the other side of the equation, they cancel out beautifully. It is a sign that you are on the right track!

Q: How do I know if I should keep the absolute value signs in the logarithm?

In JEE problems, we usually look for the domain where the solution is valid. Since our initial condition is $y(1) = 1$ (both positive), we can safely assume $x$ and $y$ are positive in the neighborhood of our point, allowing us to drop the absolute value bars without losing generality.

Q: Is it possible to solve this without the substitution?

For this specific structure, the homogeneous substitution is the most direct path. While some differential equations can be solved by inspection or by recognizing exact differentials, this one is designed to test your mastery of the homogeneous method. Stick to the process; it is the most reliable tool in your arsenal.

Q: Why do we reject the negative root for $x = \pm \sqrt{3}e$?

We look at the initial condition $y(1) = 1$. Since we are starting at $x=1$ and moving toward $y=e$, we are operating in the first quadrant where $x$ must be positive. Therefore, we discard the negative root as it falls outside the physical context of our specific solution curve.

Question

If $\frac{dy}{dx} = \frac{xy}{x^2+y^2}$; $y(1) = 1$: then a value of $x$ satisfying $y(x) = e$ is:

Options:

(A) $\sqrt{3}e$

(B) $\frac{e}{\sqrt{2}}$

(C) $\sqrt{2}e$

(D) $\frac{\sqrt{3}}{2}e$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, might look like a tangled mess of variables.

We are looking at the differential equation:

$$\frac{dy}{dx} = \frac{xy}{x^2+y^2}$$

This equation is subject to the initial condition $y(1) = 1$. It looks innocent, but beneath that simplicity lies a beautiful structure waiting to be unveiled.

### Recognizing the Pattern

Before we dive into the algebra, let us pause and observe. In the world of differential equations, the first step is always diagnosis.

Look at the numerator: $xy$ has a degree of $1+1=2$. Look at the denominator: $x^2+y^2$ also has a degree of $2$.

When the degree of every term is identical, we call this a **Homogeneous Differential Equation**. This is a powerful realization, as it tells us that the system has a scaling symmetry. If we scale $x$ and $y$ by some factor $k$, the ratio remains unchanged.

### The Substitution Strategy

To break the coupling between $x$ and $y$, we employ the classic substitution $y = vx$, where $v$ is a function of $x$. By the product rule, the derivative becomes:

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

Now, watch what happens when we substitute these into our original equation:

$$v + x \frac{dv}{dx} = \frac{x(vx)}{x^2+(vx)^2}$$

Notice how the $x^2$ terms appear in both the numerator and the denominator. We can factor them out and cancel them entirely. Suddenly, the complexity collapses into:

$$v + x \frac{dv}{dx} = \frac{v}{1+v^2}$$

### The Dance of Variables

Now, we isolate the derivative. Subtracting $v$ from both sides gives us:

$$x \frac{dv}{dx} = \frac{v - v(1+v^2)}{1+v^2}$$

This simplifies to:

$$x \frac{dv}{dx} = -\frac{v^3}{1+v^2}$$

We have successfully separated the variables. By rearranging, we get:

$$\frac{1+v^2}{v^3} dv = -\frac{1}{x} dx$$

This is where the magic happens. We split the left side into $\int (v^{-3} + v^{-1}) dv$. Integrating this is straightforward:

$$-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$$

### Returning to Reality

We must now return to our original variables. Substituting $v = \frac{y}{x}$ back into our equation, we get:

$$-\frac{x^2}{2y^2} + \ln\left|\frac{y}{x}\right| = -\ln|x| + C$$

Using the properties of logarithms, $\ln|\frac{y}{x}|$ becomes $\ln|y| - \ln|x|$. The $-\ln|x|$ terms on both sides cancel out perfectly, leaving us with the elegant particular solution:

$$-\frac{x^2}{2y^2} + \ln|y| = C$$

### The Final Stretch

Using our initial condition $y(1) = 1$, we find $C = -\frac{1}{2}$. Now, we simply need to find $x$ when $y = e$.

Substituting these values into our equation:

$$-\frac{x^2}{2e^2} + \ln|e| = -\frac{1}{2}$$

Since $\ln(e) = 1$, we have:

$$-\frac{x^2}{2e^2} + 1 = -\frac{1}{2}$$

A little bit of arithmetic leads us to $\frac{x^2}{2e^2} = \frac{3}{2}$, which simplifies to $x^2 = 3e^2$. Taking the positive root, we arrive at our destination:

**$x = \sqrt{3}e$**

You see? What began as a daunting differential equation was merely a puzzle of symmetry. By trusting the process and keeping your cool, you have navigated the complexity to find the truth.

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}$ ; $y (1) = 1$ : then a value of $x$ satisfying $y (x) = e$ is:

Select Answer:

Visualized Solution

Identify the Equation Type

The Standard Substitution

Algebraic Substitution

Simplifying the Right Hand Side

Isolating the Variable Derivative

Variable Separation

Executing the Integration

Reverting to Original Variables

Applying the Initial Condition

Solving for the Target Value

Final Calculation and Conclusion

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

Recognizing the Pattern

The Substitution Strategy

The Dance of Variables

Returning to Reality

The Final Stretch

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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 dxdy​=x2+y2xy​; y(1)=1: then a value of x satisfying y(x)=e is: