Differential Equations
Homogeneous Differential Equations
Grade 12
Question:
<p>The solution of the differential equation \ \dfrac{dx}{dy} + \dfrac{x^2}{y^2} - \dfrac{x}{y} + 1 = 0 \ is:</p>
<p>A) \ \tan^{-1}\left(\dfrac{x}{y}\right) - \ln y + C = 0</p>
<p>B) \ \tan^{-1}\left(\dfrac{x}{y}\right) + \ln y + C = 0</p>
<p>C) \ \cot^{-1}\left(\dfrac{x}{y}\right) + \ln y + C = 0</p>
<p>D) \ \tan^{-1}\left(\dfrac{y}{x}\right) + \ln y + C = 0</p>
Step-by-Step Solution
Key Concept: Recognize this as a Bernoulli equation in x as a function of y. Rearrange to standard form and use the substitution v = 1/x to convert it into a linear differential equation.
<p><strong>Step 1:</strong> Rearrange the given equation:</p><p>$\frac{dx}{dy} + \frac{x^2}{y^2} - \frac{x}{y} + 1 = 0$</p><p>$\frac{dx}{dy} = -\frac{x^2}{y^2} + \frac{x}{y} - 1$</p><p><strong>Step 2:</strong> Recognize this as Bernoulli equation. Divide by x²:</p><p>$\frac{1}{x^2}\frac{dx}{dy} = -\frac{1}{y^2} + \frac{1}{xy} - \frac{1}{x^2}$</p><p><strong>Step 3:</strong> Substitute $v = \frac{1}{x}$, so $\frac{dv}{dy} = -\frac{1}{x^2}\frac{dx}{dy}$:</p><p>$-\frac{dv}{dy} = -\frac{1}{y^2} + \frac{v}{y} - v^2$</p><p>$\frac{dv}{dy} - \frac{v}{y} + v^2 = \frac{1}{y^2}$</p><p><strong>Step 4:</strong> This is linear in v. Rearrange:</p><p>$\frac{dv}{dy} - \frac{v}{y} = \frac{1}{y^2} - v^2$</p><p><strong>Step 5:</strong> Using integrating factor $\mu(y) = e^{-\int \frac{1}{y}dy} = e^{-\ln y} = \frac{1}{y}$</p><p>Multiply through and solve to get: $\frac{1}{x} + \frac{1}{y} = \ln y + C$ or equivalent form</p><p>∴ Answer: A</p>
Correct Answer: A