Differential Equations
Differential Equations
Allen Star Batch
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 tangent line equation y - y₀ = dy/dx(X - x₀) to find intercepts A and B on axes, apply the given ratio BP:AP = 3:1 to establish a relationship between x, y, and dy/dx, then solve the resulting differential equation x(dy/dx) + 3y = 0 using separation of variables.
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