The solution of primitive integral equation (x² + y²) dy = xy dx is y = y(x). If y(1) = 1 and y(x₀)

Q: Why do we use $y = vx$ instead of $x = vy$? Does it matter?

Great question! In this specific problem, using $y = vx$ is the standard path because it directly gives us $\frac{dy}{dx}$ in terms of $v$ and $x$. You *could* use $x = vy$, which would lead to $\frac{dx}{dy} = \dots$, but it often complicates the algebra depending on the structure of the equation. Always look at the equation: if it's easier to express $\frac{dy}{dx}$ as a function of $\frac{y}{x}$, stick with $y = vx$.

Q: I got stuck during the integration of $\frac{1+v^2}{v^3}$. How do I handle that?

Don't panic! When you have a single term in the denominator like $v^3$, the best strategy is to split the fraction. Rewrite $\frac{1+v^2}{v^3}$ as $\frac{1}{v^3} + \frac{v^2}{v^3}$, which simplifies to $v^{-3} + \frac{1}{v}$. Now, you can integrate each term separately using the power rule and the natural log rule. It's all about breaking the complex into the simple!

Q: Why do we need the initial condition $y(1) = 1$?

Think of the general solution as a family of curves—an infinite set of possibilities. The initial condition acts as a 'pin' that fixes our specific curve in the coordinate plane. Without it, we have an arbitrary constant $C$. By plugging in $x=1$ and $y=1$, we lock that constant down, giving us the unique path that passes through that specific point.

Question

The solution of primitive integral equation $(x^2 + y^2) dy = xy dx$ is $y = y(x)$. If $y(1) = 1$ and $y(x_0) = e$, then $x_0$ is equal to

Options:

(A) $\sqrt{2(e^2 - 1)}$

(B) $\sqrt{2(e^2 + 1)}$

(C) $\sqrt{3} e$

(D) $\sqrt{\frac{e^2 + 1}{2}}$

Sigma Percentile · Accepted Answer

### The Symphony of Homogeneous Equations

Welcome, future engineers. Today, we are going to peel back the layers of a classic differential equation problem.

Often, when students see an equation like $(x^2 + y^2) dy = xy dx$, they freeze. They see the $x^2$, the $y^2$, and the $xy$, and they panic. But I want you to see something different; I want you to see the symmetry. In the world of differential equations, symmetry is power.

### Phase 1: The Recognition

Look at the equation again: $(x^2 + y^2) dy = xy dx$. If we rearrange this to isolate the derivative, we get:

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

Now, look at the degrees. The numerator $xy$ is a product of two variables, so its degree is $1+1=2$. The denominator $x^2 + y^2$ is a sum of two terms, each of degree 2.

When the numerator and denominator share the same degree, we are dealing with a **Homogeneous Differential Equation**. This is a beautiful structure because it tells us that the ratio of $y$ to $x$ is the key to unlocking the solution.

### Phase 2: The Bridge of Substitution

To solve this, we need a bridge. We need to transform this equation into something we can actually integrate. We use the substitution $y = vx$, where $v$ is a function of $x$.

This is the standard tool in our arsenal. But we cannot just substitute $y$; we must also substitute the derivative. Using the product rule on $y = vx$, we find that:

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

Now, watch the magic happen. When we plug these into our equation, we get:

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

Notice how the $x^2$ terms in the denominator factor out and cancel with the $x^2$ in the numerator. We are left with:

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

The complexity vanishes, leaving us with a much friendlier expression.

### Phase 3: The Calculus of Separation

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

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

With a little algebraic housekeeping, this becomes:

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

Now, we separate the variables:

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

This is where many students stumble, but you won't. Split the fraction on the left:

$$ \int (v^{-3} + v^{-1}) dv = -\int \frac{1}{x} dx $$

The integration is straightforward:

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

We are almost there. Substituting $v = \frac{y}{x}$ back into the equation, we get:

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

Using the properties of logarithms, $\ln|\frac{y}{x}| + \ln|x| = \ln|y|$. Our general solution emerges:

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

### Phase 4: The Particular Solution

We are not just looking for any curve; we are looking for the specific curve that passes through $(1, 1)$. Plugging these values into our general solution, we find:

$$ \ln(1) - \frac{1^2}{2(1)^2} = C $$

This simplifies to $0 - \frac{1}{2} = C$. Thus, $C = -\frac{1}{2}$. Our particular solution is:

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

### The Final Step

Finally, we need to find $x_0$ such that $y(x_0) = e$. Substituting $y=e$ into our equation, we get:

$$ \ln(e) - \frac{x_0^2}{2e^2} = -\frac{1}{2} $$

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

$$ 1 - \frac{x_0^2}{2e^2} = -\frac{1}{2} $$

Rearranging this gives:

$$ \frac{x_0^2}{2e^2} = 1 + \frac{1}{2} = \frac{3}{2} $$

Multiplying by $2e^2$, we get $x_0^2 = 3e^2$. Taking the square root, we arrive at the final answer:

$$ x_0 = \sqrt{3}e $$

See? It wasn't just a calculation; it was a journey. You identified the structure, built the bridge, navigated the calculus, and pinned down the specific solution. Keep this mindset, and no differential equation will ever intimidate you again.

Animated Solution for Mathematics - Differential Equations: The solution of primitive integral equation $(x^{2} + y^{2}) d y = x y d x$ is $y = y (x)$ . If $y (1) = 1$ and $y (x_{0}) = e$ , then $x_{0}$ is equal to

Select Answer:

Visualized Solution

Identifying the Equation Type

The Homogeneous Substitution

Differentiating the Substitution

Substituting into the Equation

Separating the Variables

Integrating Both Sides

Returning to Original Variables

Applying Initial Conditions

The Particular Solution

Finding the Target Value $x_{0}$

Final Calculation

Summary and Takeaways

The Sigma Insight: Homogeneous Differential Equations

The Symphony of Homogeneous Equations

Phase 1

Phase 2

Phase 3

Phase 4

The Final Step

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