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प्रश्न
उत्तर
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
\[\int\frac{1}{y}dy = \int\frac{2 x^3}{x - 1}dx\]
\[ \Rightarrow \log \left| y \right| = 2\int\frac{x^3 - 1 + 1}{x - 1}dx\]
\[ \Rightarrow \log \left| y \right| = 2\int\frac{\left( x - 1 \right)\left( x^2 + x + 1 \right) + 1}{x - 1}dx\]
\[ \Rightarrow \log \left| y \right| = 2\int\left( x^2 + x + 1 \right)dx + 2\int\frac{1}{x - 1}dx\]
\[ \Rightarrow \log \left| y \right| = \frac{2}{3} x^3 + x^2 + 2x + \log \left| x - 1 \right| + C\]
\[\text{ Hence, }\log \left| y \right| = \frac{2}{3} x^3 + x^2 + 2x + \log \left| x - 1 \right| + \text{ C is the required solution }.\]
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