Question

Solve the differential equation: $(x^{2}-yx^{2}) dy+(y^{2}+xy^{2}) dx = 0$

Answer 1

To solve the differential equation:

$(x^{2}-yx^{2}) dy+(y^{2}+xy^{2}) dx = 0$

first separate the variables than integrate both sides

$(x^{2}-yx^{2}) dy+(y^{2}+xy^{2}) dx = 0 \\ \\ (x^{2}-yx^{2}) dy = - (y^{2}+xy^{2}) dx \\ \\ {x}^{2} (1 - y)dy = - {y}^{2} (1 + x)dx \\ \\ \frac{(1 - y)}{ {y}^{2} } dy = - \frac{(1 + x)}{ {x}^{2} } dx \\ \\ \frac{1}{ {y}^{2} } dy - \frac{1}{y} dy = - \frac{1}{ {x}^{2} } dx- \frac{1}{x} dx$
integrate both sides

$\int\frac{1}{ {y}^{2} } dy - \int\frac{1}{y} dy = -\int \frac{1}{ {x}^{2} } dx- \int\frac{1}{x} dx \\ \\ - \frac{1}{y} - log \: y = \frac{1}{x} - log \: x + c \\ \\ \frac{1}{y} + log \: y = - \frac{1}{x} + log \: x - c \\ \\ \frac{1 + ylog \: y}{y} = \frac{ - 1 + x \: log \: x - cx}{x} \\ \\ x + xy \: log \: y = - y + xy \: log \: x - cxy \\ \\ y + xy(log \: y - log \: x + c) + x = 0$
is the desired solution.