Q.For the differential equation, find the particular solution $(x - y^2 x) \, dx - (y + x^2 y) \, dy = 0$ when $x = 2$, $y = 0$

Factor the equation as $x(1-y^2)\,dx = y(1+x^2)\,dy$, separate the variables, integrate to get $(1+x^2)(1-y^2) = c$, then use $x=2,\ y=0$ to obtain $(1+x^2)(1-y^2) = 5$.

Given $(x - y^2 x)\,dx - (y + x^2 y)\,dy = 0$. Factor each bracket:

$x(1 - y^2)\,dx - y(1 + x^2)\,dy = 0$

Separate the variables by dividing by $(1+x^2)(1-y^2)$:

$\dfrac{x}{1 + x^2}\,dx = \dfrac{y}{1 - y^2}\,dy$

Integrate both sides. For the left, put $1+x^2 = t \Rightarrow 2x\,dx = dt$, giving $\int \dfrac{x}{1+x^2}dx = \dfrac{1}{2}\log(1+x^2)$. For the right, put $1-y^2 = u \Rightarrow -2y\,dy = du$, giving $\int \dfrac{y}{1-y^2}dy = -\dfrac{1}{2}\log(1-y^2)$. Hence:

$\dfrac{1}{2}\log(1 + x^2) = -\dfrac{1}{2}\log(1 - y^2) + c_1$

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