Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
If a curve $y = f(x)$, passing through the point $(2, 1)$ satisfies the condition that length of subtangent is equal to slope of tangent in $1^{st}$ quadrant given that $\frac{dy}{dx} > 0$, then:
Curve $y = f(x)$ is a parabola
$y = f(x)$ is $y^2 = \frac{x}{2}$
$y = f(x)$ is $y^2 = x - 1$
Area bounded by $y = f(x)$, $y$-axis and $y = 0$, $y = 3$ is $18$ sq. units
Step-by-Step Solution
Key Concept: The length of subtangent at a point on curve y = f(x) is |y/(dy/dx)|. Setting this equal to the slope dy/dx gives y/(dy/dx) = dy/dx, which leads to (dy/dx)² = y, so dy/dx = √y (since dy/dx > 0). Solving this separable DE: 2√y = x + c, and using the point (2,1) yields the curve y² = x/2.
For orthogonal curves, $\left|\frac{dx}{dy}\right| \cdot \left|\frac{dy}{dx}\right| = 1$ implies $y\frac{dx}{dy} - \frac{dy}{dx} = 0$, giving $\frac{dy}{dx} = \sqrt{y}$ or $2\sqrt{y} = x + c$. Using the condition that the curve passes through $(2,1)$ gives $c = 0$, so $y^2 = \frac{x^2}{4}$. The area under the curve is $\int 2\sqrt{y}\,dy = 4\sqrt{3}$ square units.
Correct Answer: 1,2,4