Differential Equations
Variable Separable / Substitution
Grade 12
Question:
<p>Given the differential equation \(\dfrac{dy}{dx} = (x - y)^2\), find the general solution.</p>
<p>\(\dfrac{1}{2}\log_e\!\left(\dfrac{1+x-y}{1-x+y}\right) = x + \lambda\)</p>
<p>\(\log_e(x - y) = x + \lambda\)</p>
<p>\(\tan^{-1}(x - y) = x + \lambda\)</p>
<p>\(\dfrac{1}{2}\log_e\!\left(\dfrac{1-x+y}{1+x-y}\right) = x + \lambda\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a Clairaut-type or reducible equation by substituting v = x - y to convert it into a separable form. The substitution transforms a seemingly complex nonlinear DE into a separable equation in v.
<p><strong>Step 1:</strong> Recognize the substitution form. Let v = x - y, so y = x - v.</p><p><strong>Step 2:</strong> Differentiate: $\frac{dy}{dx} = 1 - \frac{dv}{dx}$</p><p><strong>Step 3:</strong> Substitute into the original equation: $1 - \frac{dv}{dx} = v^2$</p><p><strong>Step 4:</strong> Rearrange to separable form: $\frac{dv}{dx} = 1 - v^2 = (1-v)(1+v)$</p><p><strong>Step 5:</strong> Separate variables: $\frac{dv}{(1-v)(1+v)} = dx$</p><p><strong>Step 6:</strong> Use partial fractions: $\frac{1}{(1-v)(1+v)} = \frac{1}{2}\left(\frac{1}{1-v} + \frac{1}{1+v}\right)$</p><p><strong>Step 7:</strong> Integrate: $\frac{1}{2}\left[-\ln|1-v| + \ln|1+v|\right] = x + C$</p><p><strong>Step 8:</strong> Simplify: $\ln\left|\frac{1+v}{1-v}\right| = 2x + 2C$</p><p><strong>Step 9:</strong> Substitute back v = x - y: $\ln\left|\frac{2-y}{y}\right| = 2x + 2C$ or equivalently $\tan(x + C_1) = x - y$</p><p>∴ General solution: $(x-y) = \tan(x + C)$ where C is an arbitrary constant.</p>
Correct Answer: A