Differential Equations
First Order Separable Equations
Grade 12
Question:
<p>A curve <span>\(y = f(x)\)</span> passes through the point <span>\(P(1, 1)\)</span>. The normal to the curve at point <span>\(P\)</span> is <span>\(a(y - 1) + (x - 1) = 0\)</span>. If the slope of the tangent at any point on the curve is proportional to the ordinate at that point, then the equation of the curve is</p>
<p>(a) <span>\(y = e^{ax - 1}\)</span></p>
<p>(b) <span>\(y = e^{ax + 1}\)</span></p>
<p>(c) <span>\(y = e^{ax} + a\)</span></p>
<p>(d) <span>\(y = e^{a(x - 1)}\)</span></p>
Step-by-Step Solution
Key Concept: The differential equation dy/dx = ky describes exponential growth proportional to the ordinate. Apply initial conditions to find the constant and scale factor.
<p><strong>Step 1:</strong> From the normal equation <span>$a(y - 1) + (x - 1) = 0$</span>, the slope of the normal is <span>$-\frac{1}{a}$</span>, so the slope of the tangent at <span>$P$</span> is <span>$a$</span>.</p><p><strong>Step 2:</strong> The slope of tangent is proportional to the ordinate: <span>$\frac{dy}{dx} = ky$</span> where <span>$k$</span> is a constant.</p><p><strong>Step 3:</strong> Solving the differential equation: <span>$\frac{dy}{y} = k\,dx$</span> gives <span>$y = Ce^{kx}$</span>.</p><p><strong>Step 4:</strong> Using <span>$y(1) = 1$</span>: <span>$1 = Ce^k$</span>, so <span>$C = e^{-k}$</span>.</p><p><strong>Step 5:</strong> At <span>$P(1,1)$</span>, <span>$\frac{dy}{dx} = k \cdot 1 = a$</span>, thus <span>$k = a$</span>.</p><p><strong>Step 6:</strong> The equation is <span>$y = e^{a(x-1)}$</span>. ∴ Answer is A.</p>
Correct Answer: A