Variable Separable
Solve the separable differential equation \frac{dy}{dx} = \frac{1+y^2}{1+x^2} subject to the initial condition that y equals zero when x equals one.
Select the correct option:
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More differential equations Practice Questions
A cup of coffee cools according to Newton's law so that \frac{dT}{dt} = -k(T - T_s); identify the co...
A cup of coffee cools according to Newton's law so that \frac{dT}{dt} = -k(T - T_s); identify the co...
A population grows so that its rate of change is proportional to its current size, modeled by \frac{...
A population grows so that its rate of change is proportional to its current size, modeled by \frac{...
Solve the differential equation \frac{dy}{dx} = e^{x-y} and express the general solution relating th...
Solve the differential equation \frac{dy}{dx} = e^{x-y} and express the general solution relating th...
The solution of x dy - y dx = 0 is
The solution of x dy - y dx = 0 is
Solve the repeated-root equation \frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 0 and state the general s...
Solve the repeated-root equation \frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 0 and state the general s...
The length of the subtangent at any point of a curve is constant and equal to a, leading to a differ...
The length of the subtangent at any point of a curve is constant and equal to a, leading to a differ...
Examine the equation \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \frac{d^2y}{dx^2} and determine the d...
Examine the equation \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \frac{d^2y}{dx^2} and determine the d...
The family of curves y = Cx^2 has orthogonal trajectories that we must determine by replacing the sl...
The family of curves y = Cx^2 has orthogonal trajectories that we must determine by replacing the sl...
Solve the homogeneous equation \frac{dy}{dx} = \frac{y^2}{xy - x^2} by substituting y equal to v tim...
Solve the homogeneous equation \frac{dy}{dx} = \frac{y^2}{xy - x^2} by substituting y equal to v tim...
Find the general solution of the second-order constant-coefficient equation \frac{d^2y}{dx^2} - 5\fr...
Find the general solution of the second-order constant-coefficient equation \frac{d^2y}{dx^2} - 5\fr...
Consider the equation x\frac{dy}{dx} + 2y = x^2 with x positive, and identify the general solution a...
Consider the equation x\frac{dy}{dx} + 2y = x^2 with x positive, and identify the general solution a...
At every point on a curve the slope equals the ratio of the y-coordinate to the x-coordinate, and we...
At every point on a curve the slope equals the ratio of the y-coordinate to the x-coordinate, and we...
About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- variable separable
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
y=tan(arctanx−4π)
Separation of variables works whenever the derivative factors into a product of a function of y alone and a function of x alone, allowing each side to be integrated independently. Rearranging gives \frac{dy}{1+y^2} = \frac{dx}{1+x^2}, where each side is a standard arctangent form. Integrating both sides yields \arctan y = \arctan x + C, where C is the constant of integration determined by the initial data. Applying y = 0 at x = 1 gives \arctan 0 = \arctan 1 + C, so 0 = \frac{\pi}{4} + C, hence C = -\frac{\pi}{4}. Therefore \arctan y = \arctan x - \frac{\pi}{4}, and taking the tangent of both sides gives y = \tan\left(\arctan x - \frac{\pi}{4}\right). Option y = x - 1 is the tangent-line approximation, not the exact integral curve. Option y = \tan(\arctan x) collapses to x and ignores the constant entirely. Option y = \arctan x - \frac{\pi}{4} forgets to invert the arctangent on the left side. This matches the canonical JEE Advanced separable-equation method. As a final plausibility check, substituting x = 1 gives y = \tan(\frac{\pi}{4} - \frac{\pi}{4}) = \tan 0 = 0, recovering the initial condition exactly.
This medium difficulty mathematics question is from the chapter differential equations, covering the topic of variable separable. It appeared in the 2025 exam.
Looking for more practice? Explore all mathematics questions or browse differential equations questions on RankGuru.