Differential Equations
First Order Differential Equations
Grade 12
Question:
<p>A curve passes through the point (0,1) and the gradient at (x, y) on it is \(y(xy - 1)\). The equation of the curve is</p>
<p>(a) \(y(x - 1) = 1\)</p>
<p>(b) \(y(x + 1) = 1\)</p>
<p>(c) \(x(y + 1) = 1\)</p>
<p>(d) \(x(y - 1) = 1\)</p>
Step-by-Step Solution
Key Concept: We need to solve a differential equation where dy/dx = y(xy - 1). Separate variables by expressing this as a separable equation, then integrate both sides to find the curve equation.
<p><strong>Step 1:</strong> Set up the differential equation.</p><p>Given: dy/dx = y(xy - 1) and curve passes through (0,1).</p><p><strong>Step 2:</strong> Rearrange the differential equation.</p><p>dy/dx = xy² - y</p><p>Rearranging: dy/dx + y = xy²</p><p>Divide by y²: (1/y²)(dy/dx) + (1/y) = x</p><p><strong>Step 3:</strong> Recognize this as a Bernoulli equation or use substitution.</p><p>Let v = 1/y, so dv/dx = -(1/y²)(dy/dx)</p><p>Then: -dv/dx + v = x</p><p>Or: dv/dx - v = -x</p><p><strong>Step 4:</strong> Solve the linear differential equation.</p><p>dv/dx - v = -x is a first-order linear ODE.</p><p>Integrating factor: e^(-∫dx) = e^(-x)</p><p>Multiply both sides: e^(-x)(dv/dx) - e^(-x)v = -xe^(-x)</p><p>d/dx[e^(-x)v] = -xe^(-x)</p><p><strong>Step 5:</strong> Integrate both sides.</p><p>e^(-x)v = ∫-xe^(-x)dx = xe^(-x) + e^(-x) + C</p><p>v = x + 1 + Ce^x</p><p><strong>Step 6:</strong> Apply initial condition (0,1).</p><p>Since v = 1/y and y(0) = 1, we have v(0) = 1.</p><p>1 = 0 + 1 + C·e^0</p><p>1 = 1 + C, so C = 0</p><p><strong>Step 7:</strong> Find the curve equation.</p><p>v = x + 1</p><p>1/y = x + 1</p><p>y(x + 1) = 1</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b