Problem Statement
Solve the homogeneous differential equation:
$$(x^2 + xy)\,dy = (x^2 + y^2)\,dx$$
given that $y = 0$ when $x = 1$.
Verified Solution & Marking Scheme
Express in standard homogeneous form dy/dx
$$\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy} = \frac{1 + (y/x)^2}{1 + (y/x)}$$
Both numerator and denominator are homogeneous of degree 2. Substitute $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substitute and separate variables
$$v + x\frac{dv}{dx} = \frac{1 + v^2}{1 + v} \implies x\frac{dv}{dx} = \frac{1 + v^2 - v(1 + v)}{1 + v} = \frac{1 - v}{1 + v}$$
Separating variables:
$$\frac{1 + v}{1 - v}\,dv = \frac{1}{x}\,dx \implies \frac{-(1-v) + 2}{1 - v}\,dv = \frac{1}{x}\,dx \implies \left( -1 + \frac{2}{1 - v} \right)\,dv = \frac{dx}{x}$$
Integrate both sides
$$-v - 2\ln|1 - v| = \ln|x| + C$$
Substitute $v = \frac{y}{x}$:
$$-\frac{y}{x} - 2\ln\left| 1 - \frac{y}{x} \right| = \ln|x| + C \implies -\frac{y}{x} - 2\ln\left| \frac{x - y}{x} \right| - \ln|x| = C$$
$$-\frac{y}{x} - 2\ln|x - y| + 2\ln|x| - \ln|x| = C \implies -\frac{y}{x} - 2\ln|x - y| + \ln|x| = C$$
Apply initial boundary condition y(1) = 0
At $x = 1, y = 0$:
$$-0 - 2\ln|1 - 0| + \ln 1 = C \implies C = 0$$
Thus the particular solution is:
$$\ln\left( \frac{|x|}{(x - y)^2} \right) = \frac{y}{x} \implies \frac{x}{(x - y)^2} = e^{y/x}$$