<p>Let <strong>759.</strong> Let \(A = \int_0^2 y\,dx\) and the differential equation \(\dfrac{dy}{y+A} = dx\) with \(y(0)=1\). Find \(y(x)\).</p>
Step-by-Step Solution
Key Concept: The integral A appears both in the differential equation and as a definition involving y itself, creating a self-consistency condition. Solve the DE parametrically in terms of A, then use the boundary condition and integral definition to find the actual value of A.
<p><strong>Step 1:</strong> Solve the differential equation treating A as a parameter.</p><p>$$\frac{dy}{y+A} = dx \implies \ln|y+A| = x + C$$</p><p>Apply y(0) = 1: ln(1 + A) = C</p><p>$$y(x) = (1+A)e^x - A$$</p><p><strong>Step 2:</strong> Use the definition A = ∫₀² y dx to create a self-consistency equation.</p><p>$$A = \int_0^2 [(1+A)e^x - A]\,dx = (1+A)(e^2 - 1) - 2A$$</p><p><strong>Step 3:</strong> Solve for A.</p><p>$$A = (1+A)e^2 - (1+A) - 2A$$</p><p>$$A = e^2 + Ae^2 - 1 - A - 2A$$</p><p>$$A + A + 2A - Ae^2 = e^2 - 1$$</p><p>$$A(4 - e^2) = e^2 - 1$$</p><p>$$A = \frac{e^2 - 1}{4 - e^2}$$</p><p><strong>Step 4:</strong> Substitute back into y(x).</p><p>$$y(x) = \left(1 + \frac{e^2-1}{4-e^2}\right)e^x - \frac{e^2-1}{4-e^2}$$</p><p>$$y(x) = \frac{(4-e^2+e^2-1)e^x - (e^2-1)}{4-e^2} = \frac{3e^x - e^2 + 1}{4-e^2}$$</p>
Correct Answer: y(x) = (3e^x - e^2 + 1)/(4 - e^2)