Question

A solution curve of the equation $x y' = 2y$ passing through $(1, 4)$, also passes through

• A. $(2, 36)$
• B. $(2, 16)$
• C. $(2, 12)$
• D. $(2, 8)$

Hint:

Equations of the form $x y' = n y$ always have solutions $y = k x^n$. Here $n = 2 \implies y = k x^2$. Since $y(1) = 4 \implies k = 4 \implies y(2) = 4(2^2) = 16$.

Answer 1

The Correct Option is B

Step 1: Concept
This is a first-order separable ordinary differential equation. We separate variables $x$ and $y$, integrate both sides, solve for the general solution, and use the initial condition $(1, 4)$ to determine the constant of integration.

Step 2: Key Formulas and Approach
1. Differential equation: $x \frac{dy}{dx} = 2y$. 2. Separate variables: $\frac{1}{y} dy = \frac{2}{x} dx$. 3. Integrate: $\int \frac{1}{y} dy = 2 \int \frac{1}{x} dx$.

Step 3: Step-by-step Explanation

• Separate variables $x$ and $y$: \[ \frac{dy}{y} = \frac{2}{x} dx \]
• Integrate both sides: \[ \ln|y| = 2\ln|x| + C \] \[ \ln|y| = \ln(x^2) + C \]
• Exponentiate both sides: \[ y = k x^2 \quad (\text{where } k = e^C) \]
• Use the given point $(x, y) = (1, 4)$ to find $k$: \[ 4 = k(1)^2 \implies k = 4 \]
• The specific solution curve is: \[ y = 4 x^2 \]
• Test $x = 2$ in the solution equation: \[ y = 4(2)^2 = 4 \times 4 = 16 \]
• Therefore, the curve passes through the point $(2, 16)$.

Step 4: Final Answer
The solution curve passes through $(2, 16)$. Thus, Option (B) is correct.