Given the differential equation \(\frac{dy}{dx} = \frac{x}{y}\) and the initial condition \( y(2) = 5 \), we need to determine which of the following equations is true:1. \( x^2 - y^2 = -21 \)2. \( x^2 + y^2 = 21 \)3. \( x^2 - y^2 = 21 \)4. \( y^2 - x^2 = 11.5 \)Explanation:1. Option (1): \( x^2 - y^2 = -21 \)2. Option (2): \( x^2 + y^2 = 21 \)3. Option (3): \( x^2 - y^2 = 21 \)4. Option (4): \( y^2 - x^2 = 11.5 \)To solve this, we need to find the general solution to the differential equation and then use the initial condition to determine the specific solution. General Solution: Separate the variables and integrate both sides.\[\frac{dy}{dx} = \frac{x}{y}\]Rewriting, we get:\[y \, dy = x \, dx\]Integrating both sides, we have:\[\int y \, dy = \int x \, dx\]\[\frac{y^2}{2} = \frac{x^2}{2} + C\]Multiplying both sides by 2:\[y^2 = x^2 + 2C\]Using the initial condition \( y(2) = 5 \):\[5^2 = 2^2 + 2C\]\[25 = 4 + 2C\]\[21 = 2C\]\[C = \frac{21}{2}\]Thus, the specific solution is:\[y^2 = x^2 + 21\]This simplifies to:\[x^2 - y^2 = -21\]Therefore, the correct answer is option (1): \( x^2 - y^2 = -21 \).

Name: Given the differential equation \(\frac{dy}{dx} = \frac{x}{y}\) and t
Uploaded: 2024-03-20T06:47:08.000Z
Channel: Bartleby
Description: Given the differential equation \(\frac{dy}{dx} = \frac{x}{y}\) and the initial condition \( y(2) = 5 \), we need to determine which of the following equations is true:1. \( x^2 - y^2 = -21 \)2. \( x^2 + y^2 = 21 \)3. \( x^2 - y^2 = 21 \)4. \( y^2 -