Differential Equations
Formation and Solution of Differential Equations
Grade 12
Question:
<p>At any point <i>(x, y)</i> of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point <i>(-4, -3)</i>. The equation of the curve given that it passes through <i>(-2, 1)</i> is</p>
<p>(a) <i>y</i> + 3 = <i>x</i></p>
<p>(b) <i>(y</i> + 3) = <i>x</i>² + 4</p>
<p>(c) <i>y</i> - 3 = <i>(x</i> + 4)²</p>
<p>(d) <i>y</i> + 3 = <i>(x</i> + 4)²</p>
Step-by-Step Solution
Key Concept: Translate the geometric condition about slopes into a differential equation, then solve using separation of variables and apply the initial condition.
<p><strong>Step 1:</strong> Find the slope of the line segment passing through points <i>(x, y)</i> and <i>(-4, -3)</i>:</p><p>$$\text{slope} = \frac{y - (-3)}{x - (-4)} = \frac{y + 3}{x + 4}$$</p><p><strong>Step 2:</strong> According to the given condition, the slope of tangent is twice this slope:</p><p>$$\frac{dy}{dx} = 2\left(\frac{y + 3}{x + 4}\right)$$</p><p><strong>Step 3:</strong> Separate variables:</p><p>$$\frac{dy}{y + 3} = \frac{2\,dx}{x + 4}$$</p><p><strong>Step 4:</strong> Integrate both sides:</p><p>$$\int \frac{dy}{y + 3} = \int \frac{2\,dx}{x + 4}$$</p><p>$$\log|y + 3| = 2\log|x + 4| + \log|C|$$</p><p><strong>Step 5:</strong> Simplify:</p><p>$$\log|y + 3| = \log|x + 4|^2 + \log|C|$$</p><p>$$y + 3 = C(x + 4)^2$$</p><p><strong>Step 6:</strong> Apply the initial condition <i>y = 1</i> when <i>x = -2</i>:</p><p>$$1 + 3 = C(-2 + 4)^2$$</p><p>$$4 = 4C \Rightarrow C = 1$$</p><p><strong>Step 7:</strong> Therefore, the equation is:</p><p>$$y + 3 = (x + 4)^2$$</p><p>∴ Answer is (d).</p>
Correct Answer: D