Question

(1 − x²) dy + xy dx = xy² dx

Answer 1

We have,
\[\left( 1 - x^2 \right) dy + xy dx = x y^2 dx \]
\[ \Rightarrow \left( 1 - x^2 \right) dy = x y^2 dx - xy dx\]
\[ \Rightarrow \left( 1 - x^2 \right) dy = xy \left( y - 1 \right) dx\]
\[ \Rightarrow \frac{1}{y\left( y - 1 \right)} dy = \frac{x}{1 - x^2}dx\]
Integrating both sides, we get
\[\int\frac{1}{y\left( y - 1 \right)} dy = \int\frac{x}{1 - x^2}dx . . . . . (1)\]
Considering LHS of (1),
\[\text{ Let }\frac{1}{y\left( y - 1 \right)} = \frac{A}{y} + \frac{B}{y - 1}\]
\[ \Rightarrow 1 = A\left( y - 1 \right) + By . . . . . (2) \]
\[\text{ Substituting }y = 1\text{ in }(2), \]
\[1 = B \]
\[\text{ Substituting }y = 0\text{ in }(2), \]
\[1 = - A\]
\[ \Rightarrow A = - 1\]
\[\text{ Substituting the values of A and B in }\frac{1}{y\left( y - 1 \right)} = \frac{A}{y} + \frac{B}{y - 1}, \text{ we get }\]
\[\frac{1}{y\left( y - 1 \right)} = \frac{- 1}{y} + \frac{1}{y - 1}\]
\[ \Rightarrow \int\frac{1}{y\left( y - 1 \right)}dy = \int\frac{- 1}{y}dy + \int\frac{1}{y - 1}dy\]
\[ = - \log \left| y \right| + \log \left| y - 1 \right| + C_1 \]
Now, considering RHS of (2), we have
\[\int\frac{x}{1 - x^2}dx\]
\[\text{ Here, putting }1 - x^2 = t,\text{ we get }\]
\[ - 2x dx = dt\]
\[ \therefore \int\frac{x}{1 - x^2}dx = \frac{- 1}{2}\int\frac{1}{t}dt\]
\[ = \frac{- 1}{2}\log \left| t \right| + C_2 \]
\[ = \frac{- 1}{2}\log \left| 1 - x^2 \right| + C_2 ........\left[ \because t = 1 - x^2 \right]\]
\[\text{ Now, substituting the value of }\int\frac{1}{y\left( y - 1 \right)}dy\text{ and }\int\frac{x}{1 - x^2}dx\text{ in }(1),\text{ we get }\]
\[ - \log \left| y \right| + \log \left| y - 1 \right| + C_1 = \frac{- 1}{2}\log \left| 1 - x^2 \right| + C_2 \]
\[ \Rightarrow - \log \left| y \right| + \log \left| y - 1 \right| = - \frac{1}{2}\log \left| 1 - x^2 \right| + C \]
where
\[C = C_2 - C_1\]