Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>The solution of the differential equation \(y\,dx - (x+2y^2)dy = 0\) is \(x = f(y)\). If \(f(-1) = 1\), then \(f(1)\) is equal to</p>
<p>\(4\)</p>
<p>\(3\)</p>
<p>\(2\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: Rearrange the DE as dx/dy = (x+2y²)/y to recognize it as a linear first-order ODE in x as a function of y. This allows standard integrating factor method rather than attempting to solve for y(x).
<p><strong>Step 1:</strong> Rearrange the given equation y dx - (x+2y²)dy = 0</p><p>Dividing by y dy: dx/dy - x/y = 2y</p><p><strong>Step 2:</strong> This is a linear ODE in x: dx/dy - (1/y)x = 2y</p><p>Integrating factor: μ(y) = e^∫(-1/y)dy = e^(-ln|y|) = 1/y</p><p><strong>Step 3:</strong> Multiply both sides by 1/y:</p><p>(1/y)(dx/dy) - (1/y²)x = 2</p><p>d/dy(x/y) = 2</p><p><strong>Step 4:</strong> Integrate both sides with respect to y:</p><p>x/y = 2y + C</p><p>x = 2y² + Cy</p><p><strong>Step 5:</strong> Use initial condition f(-1) = 1, meaning when y = -1, x = 1:</p><p>1 = 2(-1)² + C(-1)</p><p>1 = 2 - C</p><p>C = 1</p><p><strong>Step 6:</strong> Therefore x = 2y² + y</p><p>At y = 1: f(1) = 2(1)² + 1 = 3</p><p>∴ Answer: B (f(1) = 3)</p>
Correct Answer: B