The solution of the differential equation dy/dx = (x+y)/x satisfying the condition y(1) = 1 is (a) y

Q: Wait, why is this called a homogeneous equation? It doesn't look like $y = x^n$.

That is a common point of confusion! In the context of differential equations, an equation $\frac{dy}{dx} = f(x, y)$ is called homogeneous if the function $f(x, y)$ can be written as a function of the ratio $\frac{y}{x}$ (or $\frac{x}{y}$). Here, $\frac{x+y}{x} = 1 + \frac{y}{x}$, which fits the definition perfectly. It is not about the degree of the polynomial, but about the scale-invariance of the slope.

Q: Do I really need the product rule for $y = vx$? Can't I just differentiate $y$?

You absolutely must use the product rule! Since $v$ is not a constant but a function of $x$ (i.e., $v = \frac{y(x)}{x}$), you are differentiating a product of two functions of $x$. If you forget the product rule and write $\frac{dy}{dx} = v$, you will miss the $x\frac{dv}{dx}$ term, which is the heart of the differential equation. Always treat $v$ as $v(x)$.

Q: What happens to the modulus sign in $\ln|x|$? Can I just ignore it?

In general, the integral of $\frac{1}{x}$ is $\ln|x| + C$. However, look at your initial condition: $y(1) = 1$. Since we are evaluating the function at $x = 1$, we are working in the domain where $x > 0$. In this domain, $|x| = x$, so we can safely drop the modulus sign. Always check the domain provided by the initial condition before removing absolute value signs.

Question

The solution of the differential equation $\frac{dy}{dx} = \frac{x+y}{x}$ satisfying the condition $y(1) = 1$ is

Options:

(A) $y = \ln x + x$

(B) $y = x \ln x + x^2$

(C) $y = xe^{(x-1)}$

(D) $y = x \ln x + x$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a mathematical landscape, looking at the differential equation:

$$\frac{dy}{dx} = \frac{x+y}{x}$$

At first glance, it might seem like a simple algebraic expression, but it is a gateway to understanding how functions evolve. Our mission is to find the specific curve $y(x)$ that satisfies this equation and passes through the point $(1, 1)$.

This is not just about solving for $y$; it is about finding the unique path that respects the geometry defined by the equation.

### The Homogeneous Insight

The first step in our journey is to simplify our perspective. When we look at the equation, we can perform a simple algebraic split:

$$\frac{dy}{dx} = \frac{x}{x} + \frac{y}{x} = 1 + \frac{y}{x}$$

This is the "Aha!" moment. The equation is now expressed entirely in terms of the ratio $\frac{y}{x}$.

This is the signature of a **Homogeneous Differential Equation**. It tells us that the slope of the tangent at any point $(x, y)$ depends only on the ratio of the coordinates, not on their absolute values.

### The Calculus Dance

To unlock this, we introduce a temporary variable, $v$, such that $y = vx$. This substitution is our bridge to a simpler world.

If $y = vx$, then by the product rule of differentiation:

$$\frac{dy}{dx} = v \cdot \frac{d}{dx}(x) + x \cdot \frac{dv}{dx} = v + x\frac{dv}{dx}$$

By substituting this into our equation, we get:

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

Notice the elegance: the $v$ terms on both sides cancel out perfectly, leaving us with $x\frac{dv}{dx} = 1$. The complexity has vanished, replaced by a beautiful, separable equation:

$$dv = \frac{1}{x} dx$$

### The Integration

Now, we integrate both sides. The integral of $dv$ is simply $v$, and the integral of $\frac{1}{x} dx$ is $\ln|x| + C$.

We have found the solution in terms of $v$, but we must return to our original variables. Substituting $v = \frac{y}{x}$ back into our result, we get:

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

Multiplying by $x$, we arrive at the general solution:

$$y = x \ln|x| + Cx$$

### The Final Touch

We are almost there. We have a family of curves, but we need the one that passes through $(1, 1)$.

Applying the initial condition $y(1) = 1$, we substitute $x=1$ and $y=1$ into our general solution:

$$1 = 1 \cdot \ln(1) + C \cdot 1$$

Since $\ln(1) = 0$, we find that $C = 1$. The specific curve we were looking for is:

**$y = x \ln x + x$**

You have successfully navigated the complexity, applied the boundary conditions, and arrived at the unique solution. This is the power of calculus: taking a seemingly chaotic equation and finding the precise, elegant path that satisfies the laws of the system.

Animated Solution for Mathematics - Differential Equations: The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

Select Answer:

Visualized Solution

Analyze the Equation

Simplify the Expression

Identify the Method

Differentiate the Substitution

Substitute into the Equation

Simplify and Separate

Integrate Both Sides

Back-Substitution

Apply Initial Condition

Final Solution

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Homogeneous Insight

The Calculus Dance

The Integration

The Final Touch

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