Differential Equations
Homogeneous Differential Equations
Grade 12
Question:
<p>The general solution of the differential equation \((y^2 - x^3)\,dx - xy\,dy = 0\) \((x \neq 0)\) is (where \(c\) is a constant of integration):</p>
<p>\(y^2 - 2x^2 + cx^3 = 0\)</p>
<p>\(y^2 + 2x^3 + cx^2 = 0\)</p>
<p>\(y^2 + 2x^2 + cx^3 = 0\)</p>
<p>\(y^2 - 2x^3 + cx^2 = 0\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a homogeneous differential equation by rewriting it in the form dy/dx, then use the substitution y = vx to separate variables and reduce it to an integrable form.
<p><strong>Step 1:</strong> Rewrite the equation in standard form.</p><p>$(y^2 - x^3)dx - xy\,dy = 0$ can be written as:<br>$xy\,dy = (y^2 - x^3)dx$<br>$\frac{dy}{dx} = \frac{y^2 - x^3}{xy}$</p><p><strong>Step 2:</strong> Verify homogeneity. Rewrite as:<br>$\frac{dy}{dx} = \frac{y}{x} - \frac{x^2}{y}$<br>This is homogeneous since both terms are degree 0 when written as functions of y/x.</p><p><strong>Step 3:</strong> Apply substitution $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.<br>$v + x\frac{dv}{dx} = v - \frac{x^2}{vx} = v - \frac{x}{v}$<br>$x\frac{dv}{dx} = -\frac{x}{v}$<br>$v\,dv = -\frac{dx}{x}$</p><p><strong>Step 4:</strong> Integrate both sides.<br>$\int v\,dv = -\int \frac{dx}{x}$<br>$\frac{v^2}{2} = -\ln|x| + C_1$<br>$v^2 = -2\ln|x| + C$ (where $C = 2C_1$)</p><p><strong>Step 5:</strong> Substitute back $v = \frac{y}{x}$.<br>$\frac{y^2}{x^2} = -2\ln|x| + C$<br>$y^2 = x^2(C - 2\ln|x|)$</p><p><strong>Or equivalently:</strong> $y^2 + 2x^2\ln|x| = Cx^2$</p><p>∴ Answer: D</p>
Correct Answer: D