Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
If $|y| = f(x)$ is solution of $\frac{d^2y}{dx^2} = \frac{x^3}{y^3}\frac{d^2x}{dy^2}$ such that $f(0) = 2$ and $y = g(x)$ is solution of $\frac{d^2y}{dx^2} + \frac{8y^3}{x^3} + \frac{d^2x}{dy^2} = 0$ such that $g(1) = \frac{1}{3}$, then:
Domain of region $f(x) \cap g(x)$ is $[-2, 2]$
Domain of region $f(x) \cap g(x)$ is $[-\sqrt{3}, \sqrt{3}]$
Range of region $f(x) \cap g(x)$ is $[0, 2]$
Range of region $f(x) \cap g(x)$ is $[0, \sqrt{3}]$
Step-by-Step Solution
Key Concept: Using the relationship d²y/dx² = (dy/dx)² · d²x/dy² to convert between differential equations in different variables, then solving the resulting first-order equations to find f(x) = √(4-x²) and g(x) = 1/√(x³+3) by applying initial conditions f(0)=2 and g(1)=1/3.
Note that $\frac{d^2y}{dx^2} = (\frac{dy}{dx})^2 \frac{d^2x}{dy^2}$, which is the relationship between second derivatives. If $y = f(x)$ is a solution of $\frac{dy}{dx} = -\frac{x}{y}$, then $y = g(x) = \frac{1}{3}x^2$ satisfies $\frac{dy}{dx} = \frac{2y}{x}$. The curves $f(x) = \sqrt{4-x^2}$ and $g(x) = \frac{1}{3}x^2$ are solutions to their respective differential equations, as shown by the parabolic region diagram.
Correct Answer: 2,3