Question

(1 + x²) dy = xy dx

Answer 1

We have,
\[\left( 1 + x^2 \right) dy = xy\ dx\]
\[ \Rightarrow \frac{1}{y}dy = \frac{x}{1 + x^2}dx\]
Integrating both sides, we get
\[\int\frac{1}{y}dy = \int\frac{x}{1 + x^2}dx\]
\[\text{ Substituting }1 + x^2 = t,\text{ we get }\]
\[2x\ dx = dt\]
\[ \therefore \int\frac{1}{y}dy = \frac{1}{2}\int\frac{1}{t}dt\]
\[ \Rightarrow \log\left| y \right| = \frac{1}{2}\log\left| t \right| + \log C \]
\[ \Rightarrow \log\left| y \right| = \frac{1}{2}\log\left| 1 + x^2 \right| + \log C .........\left(\because t = 1 + x^2\right)\]
\[ \Rightarrow \log\left| y \right| = \log\left[ C\sqrt{1 + x^2} \right]\]
\[ \Rightarrow y = C\sqrt{1 + x^2}\]
\[\text{ Hence, }y = C\sqrt{1 + x^2}\text{ is the required solution.}\]