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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

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

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

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

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

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

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

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

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

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

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