Question

Y×x×y'=(y^2+9)×(x^2+4)

Answer 1

$yxy'=(y^2+9)(x^2+4)\\\\\frac{dy}{dx}=\frac{(y^2+9)(x^2+4)}{yx}\\\\\int \frac{y\, dy}{y^2+9}=\int \frac{(x^2+4)\, dx}{x}\\\\\frac{1}{2}\int \frac{2y\, dy}{y^2+9}=\int (x+\frac{4}{x})\, dx\\\\\frac{1}{2}\cdot ln(y^2+9)=\frac{x^2}{2}+4\, ln|x|+C$