Differential Equations
Functional equations and differential equations
Grade None
Question:
<p>\(f(x) \cdot g(y) = g'(y) - f'(x) \cdot g(y)\), \(\forall\, x, y \in R\) and \(g'(0) = 1\), \(g(0) = 1\), \(f'(0) = -5\), then</p>
<p>\(f(0) = 6\)</p>
<p>\(f(1) = e\)</p>
<p>\(g(1) = e\)</p>
<p>\(g(-1) = \dfrac{1}{e}\)</p>
Step-by-Step Solution
Key Concept: Rearrange the functional equation to separate variables by treating x and y independently, then recognize that both sides must equal a constant since they depend on different variables.
<p><strong>Step 1: Rearrange the given equation</strong></p><p>Given: f(x)·g(y) = g'(y) - f'(x)·g(y)</p><p>Rearrange: f(x)·g(y) + f'(x)·g(y) = g'(y)</p><p>Factor: g(y)[f(x) + f'(x)] = g'(y)</p><p><strong>Step 2: Separate variables</strong></p><p>Since the left side depends only on x and the right side only on y, both must equal a constant k:</p><p>f(x) + f'(x) = k and g'(y) = k·g(y)</p><p><strong>Step 3: Use initial conditions to find k</strong></p><p>From g'(y) = k·g(y) with g(0) = 1, g'(0) = 1:</p><p>g'(0) = k·g(0) ⟹ 1 = k·1 ⟹ k = 1</p><p><strong>Step 4: Solve for f(x)</strong></p><p>From f(x) + f'(x) = 1 with f'(0) = -5:</p><p>f'(0) = 1 - f(0) ⟹ -5 = 1 - f(0) ⟹ f(0) = 6</p><p>Solving the differential equation f' + f = 1 gives: f(x) = 1 + Ce^(-x)</p><p>With f(0) = 6: C = 5, so f(x) = 1 + 5e^(-x)</p><p><strong>Step 5: Solution results</strong></p><p>• f(x) = 1 + 5e^(-x), g(y) = e^y satisfy all conditions ✓</p><p>• f(1) = 1 + 5e^(-1) ✓</p><p>• g'(1) = e^1 = e ✓</p><p>∴ Answer: ACD</p>
Correct Answer: ACD