Problem 69 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation of the curve satisfying the differential equation \(\sqrt{x-y \frac{d y}{d x}}=\left|x^{2}-y^{2}\right|\) and passing through the point \((1,0)\) is (A) \((2 x-1)+\frac{1}{x^{2}-y^{2}}=0\) (B) \((3 x-2)+\frac{1}{x^{2}-y^{2}}=0\) (C) \((2 x-3)+\frac{1}{x^{2}-y^{2}}=0\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is Option (C).

Step 1: Analyze the given differential equation

The given differential equation is \( \sqrt{x-y \frac{d y}{d x}}=\left|x^{2}-y^{2}\right| \). Examine the structure and manipulate it to separate variables or find a simpler form.

Step 2: Simplifying the differential equation

Square both sides of the equation to eliminate the square root.\[x - y \frac{d y}{d x} = (x^2 - y^2)^2\]

Step 3: Substitute the point into potential solutions

For each option, substitute the point \((1,0)\). Start with option (A): \(2(1)-1 + \frac{1}{1^2 - 0^2} = 0 \)This simplifies to \(1 + 1 = 2 \), which is not zero.

Step 4: Evaluate Option (B)

Substitute \((1,0)\) into (B): \(3(1) - 2 + \frac{1}{1^2 - 0^2} = 0\)This simplifies to \(1 + 1 = 2 \), which is also not zero.

Step 5: Evaluate Option (C)

Substitute \((1,0)\) into (C): \(2(1) - 3 + \frac{1}{1^2 - 0^2} = 0\)Simplifies to \(-1 + 1 = 0 \), which is zero, satisfying the condition.

Step 6: Conclusion

Since Option (C) satisfies the differential equation at the point \((1,0)\), it is the correct solution.

Key Concepts

Curve Equations

Curve equations are an integral part of calculus, particularly when dealing with differential equations. They describe a relationship between two variables, typically denoted as \(x\) and \(y\), in a geometric manner. These equations help us understand how variables change in relation to each other over a curve.A curve equation formulated from a differential equation outlines the path or trajectory taken when a specific mathematical condition is satisfied. In the given exercise, the equation of the curve has to satisfy the differential equation \(\sqrt{x-y \frac{d y}{d x}}=\left|x^{2}-y^{2}\right|\). When we find the right curve equation that meets the initial condition, it helps us accurately chart the path of change.The correct selection of the curve equation is validated by checking if the equation holds true when a specific point, such as \((1, 0)\), is substituted into it. This process ensures the pathway or the nature of change is aligned with the given conditions. By using the approved curve equation, one can determine how transformations occur across varied points on the curve.

Variable Separation

Variable separation is a technique used to simplify differential equations by isolating one variable on each side of the equation. This method is particularly effective when an equation is initially too complex to solve directly and needs simplification.In the original step-by-step solution, an attempt is made to simplify the differential equation \(\sqrt{x-y \frac{d y}{d x}}=\left|x^{2}-y^{2}\right|\), by squaring both sides. This act of simplification aimed at providing clarity and made it easier to work with the components separately. Even though this problem doesn’t directly apply a classic separation of variables, understanding it as a tool helps greatly. It aids in deriving solutions from complex differential equations by downsizing it into manageable parts.Variable separation not only helps in solving the differential equation but also elucidates the relationship between derivatives. When this technique is successfully applied, it becomes possible to integrate the separated components separately, allowing the curve equations to emerge with greater ease.

Initial Conditions

Initial conditions are specific values used to solve for constants in differential equations, ensuring that the solution captures the specific scenario described by the condition. They play a pivotal role in ensuring that the derived solution is unique to the scenario.In the context of the given exercise, the initial condition is given as the point \((1,0)\). This means that whatever solution is found must yield a result consistent with this known value when substituted back into the equation.When evaluating which curve equation correctly satisfies the condition, we substitute the initial condition into each potential equation to verify its validity. For instance, substituting \((1,0)\) into the options provided confirmed that option \(C: (2x-3)+\frac{1}{x^2-y^2} = 0\) meets the initial condition.Using initial conditions in this manner ensures that the solution is not just theoretically correct, but also accurately mocks the real-world scenario or path described by the differential equation.