Exams › JEE Advanced › Maths

By writing the equation as the exact differential of an expression in x and y, find the general solution of (2x³ - x y²) dx + (2y³ - x² y) dy = 0.

1. x⁴ - x² y² + y⁴ = c
2. x⁴ + x² y² - y⁴ = c
3. x⁴ - x² y² - y⁴ = c
4. x⁴ + x² y² + y⁴ = c

Correct answer: x⁴ - x² y² + y⁴ = c

Solution

The equation becomes d(x⁴/2) + d(y⁴/2) - d(x² y² /2) = 0, integrating to x⁴ + y⁴ - x² y² = c.

Related JEE Advanced Maths questions

• Consider the differential equation associated with y = Σ (from i=1 to 3) C_i e^(m_i x), where C_i represents arbitrary constants and m₁, m₂, m₃ are the solutions of m³ - 7m + 6 = 0. If the equation is expressed as d³y/dx³ - 7 dy/dx + k y = 0, determine the value of k.
• Determine the order of the highest derivative raised to a power in the equation: (dy/dx)⁴ - 2x (d³y/dx³)² x² d²y/dx² d³y/dx³.
• The provided differential equation is expressed as a polynomial involving derivatives such as d²y/dx² and d³y/dx³. Among these, d³y/dx³ is the derivative of the highest order, and its greatest power in the equation is 2. What is the degree of this differential equation?
• When solving the differential equation d²y/dx² = 6x - 4, the integration gives dy/dx = 3x² - 4x + A. If dy/dx becomes zero at x = 1, what is the value of A?
• A curve passes through the point (1, π/6). Let the slope of the curve at each point (x, y) be y/x + sec(y/x), x > 0. Then the equation of the curve is -
• Consider y(x) as a solution to the differential equation (1 + e^x) dy/dx + y e^x = 1, with the initial condition y(0) = 2. Which of the following assertions is/are accurate?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →