The given differential equation is: \[ xdy - ydx = 0 \] Rearranging terms: \[ \frac{dy}{dx} = \frac{y}{x} \] This is a separable differential equation. By separating variables: \[ \frac{dy}{y} = \frac{dx}{x} \] Integrating both sides: \[ \ln|y| = \ln|x| + C \] This implies: \[ y = Cx \]
So, the correct answer is (C) : Straight line passing through origin.
We are given the differential equation:
\[ x\,dy - y\,dx = 0 \]
Step 1: Rearranging the equation
Divide both sides by \( dx \): \[ x \frac{dy}{dx} - y = 0 \Rightarrow \frac{dy}{dx} = \frac{y}{x} \]
Step 2: Solve the differential equation
This is a separable differential equation: \[ \frac{dy}{y} = \frac{dx}{x} \] Integrating both sides: \[ \int \frac{dy}{y} = \int \frac{dx}{x} \Rightarrow \ln |y| = \ln |x| + C \] Taking exponentials on both sides: \[ |y| = A |x| \Rightarrow y = Cx \quad \text{(where } C \text{ is a constant)} \]
Conclusion: The solution represents a family of straight lines passing through the origin.
Final Answer: Straight line passing through origin
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \). If \( f'(0) = 4a \) and \( f \) satisfies \( f''(x) - 3a f'(x) - f(x) = 0 \), where \( a > 0 \), then the area of the region R = {(x, y) | 0 \(\leq\) y \(\leq\) f(ax), 0 \(\leq\) x \(\leq\) 2 is :
In a practical examination, the following pedigree chart was given as a spotter for identification. The students identify the given pedigree chart as 