Differential Equations
Variable substitution in ODEs
Grade 12
Question:
<p>The solution of the differential equation \(\dfrac{dy}{dx} = (x - y)^2\), when \(y(1) = 1\), is:</p>
<p>\(\log_e\left|\dfrac{2-x}{2-y}\right| = x - y\)</p>
<p>\(-\log_e\left|\dfrac{1-x+y}{1+x-y}\right| = 2(x-1)\)</p>
<p>\(-\log_e\left|\dfrac{1+x-y}{1-x+y}\right| = x + y - 2\)</p>
<p>\(\log_e\left|\dfrac{2-y}{2-x}\right| = 2(y-1)\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a Clairaut-type or substitution problem. Use the substitution v = x - y to transform it into a separable equation, which converts the nonlinear ODE into a tractable form.
<p><strong>Step 1:</strong> Substitute v = x - y, so dv/dx = 1 - dy/dx</p><p>This gives: dy/dx = 1 - dv/dx</p><p><strong>Step 2:</strong> Substitute into the original equation: 1 - dv/dx = v²</p><p>Rearranging: dv/dx = 1 - v²</p><p><strong>Step 3:</strong> Separate variables: dv/(1 - v²) = dx</p><p><strong>Step 4:</strong> Use partial fractions: 1/(1-v²) = 1/2[1/(1-v) + 1/(1+v)]</p><p>Integrating: (1/2)ln|(1+v)/(1-v)| = x + C</p><p><strong>Step 5:</strong> Apply initial condition y(1) = 1, so v(1) = 1 - 1 = 0</p><p>At x = 1, v = 0: (1/2)ln(1) = 1 + C, so C = -1</p><p><strong>Step 6:</strong> Solve for v: (1/2)ln|(1+v)/(1-v)| = x - 1</p><p>Back-substitute v = x - y to obtain the implicit solution or explicit form depending on answer choices.</p><p>∴ Answer: B</p>
Correct Answer: B