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( X − 1 ) D Y D X = 2 X 3 Y

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प्रश्न

\[\left( x - 1 \right)\frac{dy}{dx} = 2 x^3 y\]
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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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अध्याय 21: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 21 Differential Equations
Exercise 22.07 | Q 4 | पृष्ठ ५५

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