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The given equation is $(x^2 + y^2)dy = xydx$. We can rewrite this as: $$ \frac{dy}{dx} = \frac{xy}{x^2 + y^2} $$ This is a homogeneous differential equation.
Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the equation: $$ v + x\frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2(1 + v^2)} = \frac{v}{1 + v^2} $$
Rearranging the terms: $$ x\frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v - v^3}{1 + v^2} = \frac{-v^3}{1 + v^2} $$ Separating variables: $$ \frac{1 + v^2}{v^3} dv = -\frac{1}{x} dx $$ $$ (v^{-3} + v^{-1}) dv = -\frac{1}{x} dx $$
Integrating both sides: $$ \int (v^{-3} + v^{-1}) dv = -\int \frac{1}{x} dx $$ $$ \frac{v^{-2}}{-2} + \ln|v| = -\ln|x| + C $$ $$ -\frac{1}{2v^2} + \ln|v| + \ln|x| = C $$ $$ -\frac{1}{2v^2} + \ln|vx| = C $$ Since $y = vx$, we have: $$ -\frac{x^2}{2y^2} + \ln|y| = C $$
The condition $y(0) = 1$ implies $x=0, y=1$. However, substituting $x=0$ into the equation results in an undefined term $-\frac{0}{2(1)^2} + \ln(1) = C$, which gives $C=0$. Thus, the solution is: $$ \ln|y| = \frac{x^2}{2y^2} $$
Final Answer: 2y^2 \ln|y| = x^2
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