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Find a solution for the equation $x \frac{d y}{d x}+y^{2}=x^{2} y^{2} \ln (x)$ where $x=1, y=-1$

$x \frac{d y}{d x}+y^{2}=x^{2} y^{2} \ln (x) \quad \div d x$

$x d y+y^{2} d x=x^{2} y^{2} \ln (x) d x$

$x d y=x^{2} y^{2} \ln (x) d x-y^{2} d x$

$x d y=y^{2}\left(x^{2} \ln (x) d x-d x\right)$

$x d y=y^{2}\left(x^{2} \ln (x)-1\right) d x$

$\div x y^{2}$

$\frac{d y}{y^{2}}={(\frac{x^{2} \ln (x)}{x}}-\frac{1}{x}) d x$

$\frac{d y}{y^{2}}=\left(x \ln (x)-\frac{1}{x}\right) d x$ integrate

$\int {y}^{-2} d y=\int\left(x \ln (x)-\frac{1}{x}\right) d x$

$u=\ln (x) \quad d u=\frac{1}{x}$
$v=\frac{x^{2}}{2} \quad d u=x$

$u \cdot v-\int v d u$

$=\frac{x^{2}}{2}\cdot \ln (x)-\int \frac{x^{2}}{2} \cdot \frac{1}{x} d x$

$=\frac{x^{2}}{2} \ln (x)-\int \frac{x}{2} d x$

$\frac{y^{-1}}{-1}=\frac{x^{2}}{2} \ln (x)-\int \frac{x}{2} d x-\ln x+c$

$-\frac{1}{y}=\frac{x^{2}}{2} \ln (x)-\frac{1}{4} x^{2}-\ln (x)+c$

$-\frac{1}{-1}=\frac{(1)^{2}}{2} \ln (1)-\frac{1}{4}(1)^{2}-\ln (1)+c$

$1=-\frac{1}{4}+c \rightarrow c=1+\frac{1}{4}=\frac{5}{4}$