Differential Equations
Linear Differential Equation
Grade 12
Question:
<p>On applying Newton–Leibniz rule to the equation<br/>\[ x[y(x) - 0] + \int_1^x y(t)\,dt = \int_1^x ty(t)\,dt + (x+1)(xy(x)-0) \]<br/>and simplifying, the solution \( y(x) \) is found to be:</p>
<p>\( y(x) = cx^3 e^{1/x} \)</p>
<p>\( y(x) = \dfrac{c}{x^3}e^{-1/x} \)</p>
<p>\( y(x) = cx^3 e^{-1/x} \)</p>
<p>\( y(x) = \dfrac{c}{x^3}e^{1/x} \)</p>
Step-by-Step Solution
Key Concept: Differentiate both sides using the Newton-Leibniz rule (Fundamental Theorem of Calculus) to convert the integral equation into a differential equation, then solve the resulting ODE.
<p><strong>Step 1:</strong> Apply Newton-Leibniz rule (differentiate both sides with respect to x):</p><p>Left side: d/dx[xy(x) + ∫₁ˣ y(t)dt] = y(x) + xy'(x) + y(x) = 2y(x) + xy'(x)</p><p><strong>Step 2:</strong> Right side: d/dx[∫₁ˣ ty(t)dt + (x+1)(xy(x))] = xy(x) + (x+1)[y(x) + xy'(x)]</p><p>= xy(x) + (x+1)y(x) + (x+1)xy'(x)</p><p><strong>Step 3:</strong> Equate and simplify:</p><p>2y(x) + xy'(x) = xy(x) + (x+1)y(x) + (x+1)xy'(x)</p><p>2y - xy - (x+1)y = (x+1)xy' - xy'</p><p>y(2 - x - x - 1) = xy'(x + 1 - 1)</p><p>-y(2x - 1) = x²y'</p><p><strong>Step 4:</strong> Separate variables: dy/y = -(2x-1)/x² dx</p><p>Integrate: ln|y| = -2ln|x| + 1/x + C</p><p>∴ y(x) = Ae^(1/x)/x² or y(x) = K·e^(1/x)·x⁻²</p>
Correct Answer: B