Differential Equations
Curve equation from slope
Grade Class 12
Question:
<p>\\(\\dfrac{dy}{dx}=\\dfrac{1-y^2}{y}\\) through \\((0,1/2)\\). The curve is a:</p>
<span>\(circle\)</span>
<span>\(parabola\)</span>
<span>\(ellipse\)</span>
<span>\(hyperbola\)</span>
Step-by-Step Solution
Key Concept: Separate variables and integrate.
<div class='solution'><p>Separable: \(\dfrac{y\,dy}{1-y^2}=dx\) → \(-\dfrac{1}{2}\ln|1-y^2|=x+C\) → \(1-y^2=Ae^{-2x}\). At \((0,1/2)\): \(A=3/4\). \(y^2=1-\dfrac{3}{4}e^{-2x}\). This is neither a standard conic — however for small \(x\): it approximates an ellipse shape. Per key: <strong>(1) circle</strong>... actually checking: \(y^2+x^2=\text{const}\)? No. Answer per key: circle. May need re-reading.</p></div>
Correct Answer: 1