Question

Solve the differential equation (x² − yx²) dy + (y² + x²y²) dx = 0, given that y = 1, when x = 1.

Answer 1

Given: ​(x² − yx²) dy + (y² + x²y²) dx = 0
Dividing both the sides by 

\[dx\],  we get:

\[\left( x^2 - y x^2 \right)\frac{dy}{dx} + \left( y^2 + x^2 y^2 \right) = 0\]

\[\Rightarrow x^2 \left( 1 - y \right)\frac{dy}{dx} + y^2 \left( 1 + x^2 \right) = 0\]

\[ \Rightarrow - x^2 \left( 1 - y \right)\frac{dy}{dx} = y^2 \left( 1 + x^2 \right)\]

\[ \Rightarrow x^2 \left( y - 1 \right)\frac{dy}{dx} = y^2 \left( 1 + x^2 \right)\]

\[ \Rightarrow \frac{\left( y - 1 \right)}{y^2}dy = \frac{1 + x^2}{x^2}dx\]

Integration both the sides:

\[\int\frac{\left( y - 1 \right)}{y^2}dy = \int\frac{1 + x^2}{x^2}dx\]

\[\frac{1}{2}\int\frac{2y}{y^2}dy - \int\frac{1}{y^2}dy = \int\frac{1}{x^2}dx + \int1 . dx\]

\[\text { Put } y^2 = t\]

\[\text { Differentiating w . r . t  }t , 2ydy = dt\]

\[ \Rightarrow \frac{1}{2}\int\frac{dt}{t} + \frac{1}{y} = - \frac{1}{x} + x\]

\[ \Rightarrow \frac{1}{2}\log\left| y^2 \right| + \frac{1}{y} = - \frac{1}{x} + x + C\]

Given: y=1, x=1

\[ \Rightarrow \frac{1}{2}\log\left| 1 \right| + 1 = - 1 + 1 + C\]

\[ \Rightarrow C = 1\]

\[\Rightarrow \frac{1}{2}\log\left| y^2 \right| + \frac{1}{y} = - \frac{1}{x} + x + 1\] is the required solution.