Question

\[\left( x - 1 \right)\frac{dy}{dx} = 2 x^3 y\]

Answer 1

We have, 
\[\left( x - 1 \right)\frac{dy}{dx} = 2 x^3 y\]
\[ \Rightarrow \frac{1}{y}dy = \frac{2 x^3}{x - 1}dx\]
Integrating both sides, we get 
\[\int\frac{1}{y}dy = \int\frac{2 x^3}{x - 1}dx\]
\[ \Rightarrow \log \left| y \right| = 2\int\frac{x^3 - 1 + 1}{x - 1}dx\]
\[ \Rightarrow \log \left| y \right| = 2\left[ \int\frac{x^3 - 1}{x - 1}dx + \int\frac{1}{x - 1}dx \right]\]
\[ \Rightarrow \log \left| y \right| = 2\left[ \int\frac{\left( x - 1 \right)\left( x^2 + x + 1 \right)}{x - 1}dx + \int\frac{1}{x - 1}dx \right]\]
\[ \Rightarrow \log \left| y \right| = 2\left[ \int\left( x^2 + x + 1 \right) dx + \int\frac{1}{x - 1}dx \right]\]
\[ \Rightarrow \log \left| y \right| = 2 \left[ \frac{x^3}{3} + \frac{x^2}{2} + x + \log \left| x - 1 \right| \right] + C\]
\[ \Rightarrow \log \left| y \right| = \frac{2}{3} x^3 + x^2 + 2x + \log \left| x - 1 \right|^2 + C\]
\[ \Rightarrow y = e^{\frac{2}{3} x^3 + x^2 + 2x + \log \left| x - 1 \right|^2 + C}\]
\[ \Rightarrow y = e^C \times e^{{log} \left| x - 1 \right|^2} \times e^{\frac{2}{3} x^3 + x^2 + 2x}\]
\[ \Rightarrow y = C_1 \left| x - 1 \right|^2 e^{\frac{2}{3} x^3 + x^2 + 2x} ..........\left[ \because e^{ln\ x} = x\text{ and where, }C_1 = e^C \right]\]
\[ \therefore y = C_1 \left| x - 1 \right|^2 e^{\frac{2}{3} x^3 + x^2 + 2x} \text{ is required solution.} \]