Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>The solution of the differential equation \(\dfrac{dy}{dx} = \dfrac{x+y}{x}\) satisfying the condition \(y(1) = 1\) is</p>
<p>\(y = \ln x + x\)</p>
<p>\(y = x\ln x + x^2\)</p>
<p>\(y = xe^{(x-1)}\)</p>
<p>\(y = x\ln x + x\)</p>
Step-by-Step Solution
Key Concept: Rewrite the equation as dy/dx = 1 + y/x to recognize it as a linear first-order ODE in standard form, then use integrating factor method with μ(x) = 1/x.
<p><strong>Step 1:</strong> Rewrite the equation in standard form:</p><p>dy/dx = (x+y)/x = 1 + y/x</p><p>Rearranging: dy/dx - y/x = 1</p><p><strong>Step 2:</strong> Identify this as a linear ODE of form dy/dx + P(x)y = Q(x), where P(x) = -1/x and Q(x) = 1</p><p><strong>Step 3:</strong> Find integrating factor: μ(x) = e^(∫P(x)dx) = e^(-∫1/x dx) = e^(-ln x) = 1/x</p><p><strong>Step 4:</strong> Multiply both sides by μ(x) = 1/x:</p><p>(1/x)dy/dx - y/x² = 1/x</p><p>This gives: d/dx(y/x) = 1/x</p><p><strong>Step 5:</strong> Integrate both sides:</p><p>y/x = ∫(1/x)dx = ln|x| + C</p><p>Therefore: y = x(ln|x| + C) = x ln x + Cx</p><p><strong>Step 6:</strong> Apply initial condition y(1) = 1:</p><p>1 = 1·ln(1) + C·1 = 0 + C</p><p>So C = 1</p><p><strong>Step 7:</strong> Final solution: y = x ln x + x = x(ln x + 1)</p><p>∴ Answer: D</p>
Correct Answer: D