Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
A tangent drawn to the curve $y = f(x)$ at P(x, y) cuts the x-axis and y-axis at A and B respectively such that BP: AP = 3:1, given that $f(1) = 1$, then:
Equation of curve is x\frac{dy}{dx} - 3y = 0
Normal at (1,1) is x + 3y = 4
Curve passes through (2, 1/8)
Equation of curve is x\frac{dy}{dx} + 3y = 0
Step-by-Step Solution
Key Concept: Use the geometric property of the subtangent and convert to a separable differential equation in terms of $\ln x$ and $\ln y$.
Given $y - y = \frac{dy}{dx}(X - x)$, the subtangent is $3(x - y\frac{dx}{dy}) + 0 = x$. This gives $x\frac{dy}{dx} + 3y = 0$ and $\frac{dy}{y} + 3\frac{dx}{x} = 0$. Integrating yields $\ln y + 3\ln x = \ln c$, or $yx^3 = c = 1$ using $f(1) = 1$.
Correct Answer: 3,4