Differential Equations
Formation of Differential Equations
Grade None
Question:
<p>The differential equation whose solution is \(Ax^2 + By^2 = 1\), where \(A\) and \(B\) are arbitrary constants is of</p>
<p>second order and second degree.</p>
<p>first order and second degree.</p>
<p>first order and first degree.</p>
<p>second order and first degree.</p>
Step-by-Step Solution
Key Concept: Eliminate two arbitrary constants A and B by differentiating twice and forming relationships between the derivatives. Each differentiation reduces the number of arbitrary constants by one.
<p><strong>Step 1:</strong> Start with the given family of curves: <em>Ax</em>² + <em>By</em>² = 1</p><p><strong>Step 2:</strong> Differentiate once with respect to <em>x</em>:<br/>2<em>Ax</em> + 2<em>By</em>(d<em>y</em>/d<em>x</em>) = 0<br/>⟹ <em>Ax</em> + <em>By</em>p = 0, where p = d<em>y</em>/d<em>x</em> ... (1)</p><p><strong>Step 3:</strong> Differentiate equation (1) again with respect to <em>x</em>:<br/><em>A</em> + <em>B</em>[p² + <em>y</em>(d²<em>y</em>/d<em>x</em>²)] = 0<br/>⟹ <em>A</em> + <em>B</em>(p² + <em>yq</em>) = 0, where q = d²<em>y</em>/d<em>x</em>² ... (2)</p><p><strong>Step 4:</strong> From equation (1): <em>A</em> = -<em>Byp</em>/x<br/>Substitute into equation (2):<br/>-<em>Byp</em>/x + <em>B</em>(p² + <em>yq</em>) = 0<br/>⟹ <em>B</em>[-<em>yp</em>/x + p² + <em>yq</em>] = 0<br/>⟹ <em>xyq</em> + <em>xp</em>² - <em>yp</em> = 0</p><p><strong>Step 5:</strong> This is a <strong>second-order differential equation</strong> (contains d²<em>y</em>/d<em>x</em>²) with no arbitrary constants.</p><p>∴ Answer: <strong>D</strong> (Second order)</p>
Correct Answer: D