The function \(y=e^{kr}\) satisfies \((\frac{d^2y}{dx^2}+\frac{dy}{dx})(\frac{dy}{dx}-y)=y\frac{dy}{dx}\). It is valid for
- exactly one value of k
- two distinct values of k
- three distinct values of k
- infinitely many values of k
The Correct Option is B
Solution and Explanation
This simplifies to: \(k^2 - 1 = (\frac{d^2x}{dy^2})\)
Now, let's analyze the resulting expression \(k^2 - 1 = (\frac{d^2x}{dy^2})\) :
Valid Domain for k: The equation \(k^2 - 1 = (\frac{d^2x}{dy^2})\) is valid for all real values of k. There are no restrictions on k based on this equation.
Valid Domain for \((\frac{d^2x}{dy^2})\): Since \((\frac{d^2x}{dy^2})\) represents the second derivative of x with respect to y, this term is valid as long as x is a function that can be differentiated twice with respect to y. There are no specific restrictions on the domain of x based on this equation.
Therefore, the differential equation is valid for all real values of k and for functions x that are twice differentiable with respect to y.
The correct option is (B): two distinct values of k.
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Concepts Used:
Second-Order Derivative
The Second-Order Derivative is the derivative of the first-order derivative of the stated (given) function. For instance, acceleration is the second-order derivative of the distance covered with regard to time and tells us the rate of change of velocity.
As well as the first-order derivative tells us about the slope of the tangent line to the graph of the given function, the second-order derivative explains the shape of the graph and its concavity.
The second-order derivative is shown using \(f’’(x)\text{ or }\frac{d^2y}{dx^2}\).
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