If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,

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

Great question! We choose $y = vx$ because the equation is already easily rearranged into the form $\frac{dy}{dx} = F(x, y)$, where $F$ is a homogeneous function of degree 0. This makes $y = vx$ the most direct path. If we had used $x = vy$, we would be solving for $\frac{dx}{dy}$, which is possible but unnecessary here.

Q: What if the degrees of the terms weren't all the same?

If the degrees were not identical, the equation would not be homogeneous, and the $y = vx$ substitution would not simplify the equation into a separable one. In that case, you would need to look for other methods, such as checking if it's a linear differential equation or an exact differential equation.

Q: Why is the constant $C$ important in the final step?

The constant $C$ represents the family of all possible solution curves. Because the problem specifies that the curve passes through the point $(1, 2)$, we use this initial condition to 'pin down' the specific curve from that family. Without finding $C$, we wouldn't have the unique function $f(x)$ needed to evaluate $f(1/2)$.

Question

If a curve $y = f(x)$, passing through the point $(1, 2)$, is the solution of the differential equation, $2x^2dy = (2xy + y^2) dx$, then $f \left(\frac{1}{2}\right)$ is equal to :

Options:

(A) $\frac{1}{1+\log_e 2}$

(B) $\frac{-1}{1+\log_e 2}$

(C) $1 + \log_e 2$

(D) $\frac{1}{1-\log_e 2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, winding path—a differential equation that seems to twist and turn in ways that defy simple integration. Today, we are looking at the equation $2x^2 dy = (2xy + y^2) dx$.

Our first step is to act like detectives. We rearrange the equation to isolate the derivative:

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

Now, look closely at the structure. The numerator has terms $2xy$ and $y^2$, both of which have a degree of 2. The denominator, $2x^2$, also has a degree of 2.

This is not a coincidence; it is the hallmark of a **homogeneous differential equation**. This symmetry is our golden ticket.

### The Transformation: Changing Our Perspective

When we encounter a homogeneous equation, we don't fight it head-on. Instead, we change our perspective. We use the substitution $y = vx$, where $v$ is a function of $x$.

To make this work, we must also transform our derivative. Using the product rule on $y = vx$, we get:

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

Now, we substitute these into our original equation:

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

Simplifying the right side, we get:

$$ \frac{2vx^2 + v^2x^2}{2x^2} $$

Notice how the $x^2$ terms cancel out beautifully? We are left with:

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

This further simplifies to:

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

Subtracting $v$ from both sides gives us the elegant result:

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

### The Calculus of Separation

We have arrived at the heart of the problem. The equation is now in **variable separable form**:

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

We integrate both sides:

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

The integral of $v^{-2}$ is $-v^{-1}$, so the left side becomes $-\frac{2}{v}$. The right side is simply $\ln|x| + C$. Substituting back $v = \frac{y}{x}$, we get:

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

This is our general solution, the map of all possible curves that satisfy the differential equation.

### The Final Reveal

We know the curve passes through $(1, 2)$, so we plug in $x = 1$ and $y = 2$ to find $C$:

$$ -\frac{2(1)}{2} = \ln|1| + C \Rightarrow -1 = 0 + C \Rightarrow C = -1 $$

Our particular solution is:

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

Finally, to find $f\left(\frac{1}{2}\right)$, we set $x = \frac{1}{2}$:

$$ -\frac{2(1/2)}{y} = \ln\left(\frac{1}{2}\right) - 1 $$

This simplifies to:

$$ -\frac{1}{y} = -\ln 2 - 1 $$

Multiplying by $-1$, we get:

$$ \frac{1}{y} = 1 + \ln 2 $$

Taking the reciprocal, we find our final answer:

$$ **y = \frac{1}{1 + \ln 2}** $$

Animated Solution for Mathematics - Differential Equations: If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :

Select Answer:

Visualized Solution

The Given Differential Equation

Identifying the Equation Type

The Substitution $y = v x$

Substituting into the Equation

Simplifying the Expression

Isolating the $v$ Terms

Separating the Variables

Integrating Both Sides

Substituting Back $v = \frac{y}{x}$

Finding the Constant $C$

The Particular Solution

Evaluating at $x = \frac{1}{2}$

Final Answer Calculation

Summary and Takeaway

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Transformation

The Calculus of Separation

The Final Reveal

Similar Questions

If the solution curve, of the differential equation $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$ passing through the point $(2, 1)$ is $tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} lo g_{e} (α + (\frac{y - 1}{x - 1})^{2}) = lo g_{e} ∣ x - 1∣$ , then $5 β + α$ is equal to

If $y = y (x)$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that $y (e) = \frac{e}{3}$ , then $y (1)$ is equal to

The solution curve of the differential equation $y \frac{d x}{d y} = x (lo g_{e} x - lo g_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x ^{2} + 3 y ^{2}}{3 x ^{2} + y ^{2}}), y (1) = 0$ is

Let $y = y (x)$ be the solution of the differential equation $(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1$ . Then $6 y^{2} (e)$ is equal to

If $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , then the solution of the equation is

Let the solution curve $y = y (x)$ of the differential equation, $[\frac{x}{x ^{2} - y ^{2}} + e^{y / x}] x \frac{d y}{d x} = x + [\frac{x}{x ^{2} - y ^{2}} + e^{y / x}] y$ pass through the points $(1, 0)$ and $(2 α, α), α > 0$ . Then $α$ is equal to

The curve amongst the family of curves, represented by the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ which passes through $(1, 1)$ is :

Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to

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

The Given Differential Equation

Identifying the Equation Type

The Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=vx

Substituting into the Equation

Simplifying the Expression

Isolating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v Terms

Separating the Variables

Integrating Both Sides

Substituting Back .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=xy​

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C