<p>If \(y(t)\) is a solution of \((1 + t)\frac{dy}{dt} - ty = 1\) and \(y(0) = -1\), then \(y(1)\) = _____.</p>
Step-by-Step Solution
Key Concept: Rewrite as a linear first-order ODE in standard form dy/dt + P(t)y = Q(t), then use the integrating factor method μ(t) = e^∫P(t)dt to solve systematically.
<p><strong>Step 1:</strong> Rewrite in standard form. Divide by (1+t):</p><p>dy/dt - [t/(1+t)]y = 1/(1+t)</p><p><strong>Step 2:</strong> Find integrating factor. P(t) = -t/(1+t), so:</p><p>∫P(t)dt = ∫[-t/(1+t)]dt = ∫[(-1 + 1/(1+t))]dt = -t + ln(1+t)</p><p>μ(t) = e^(-t + ln(1+t)) = (1+t)e^(-t)</p><p><strong>Step 3:</strong> Multiply both sides by μ(t):</p><p>(1+t)e^(-t)·dy/dt - te^(-t)y = e^(-t)</p><p>d/dt[(1+t)e^(-t)y] = e^(-t)</p><p><strong>Step 4:</strong> Integrate both sides:</p><p>(1+t)e^(-t)y = ∫e^(-t)dt = -e^(-t) + C</p><p><strong>Step 5:</strong> Apply initial condition y(0) = -1:</p><p>(1+0)e^0·(-1) = -e^0 + C</p><p>-1 = -1 + C ⟹ C = 0</p><p><strong>Step 6:</strong> Solve for y:</p><p>(1+t)e^(-t)y = -e^(-t)</p><p>y = -1/(1+t)</p><p><strong>Step 7:</strong> Find y(1):</p><p>y(1) = -1/(1+1) = -1/2</p><p>∴ Answer: <strong>-1/2</strong></p>
Correct Answer: -1