First slide

Types of solutions of differential equations

Question

$Solution of the equation x d y - [y + x y^{3} (1 + \log x)] d x = 0 is$

Moderate

A

$\frac{- x^{2}}{y^{2}} = \frac{2 x^{3}}{3} (\frac{2}{3} + \log x) + c$
B

$\frac{x^{2}}{y^{2}} = \frac{2 x^{3}}{3} (\frac{2}{3} + \log x) + c$
C

$\frac{- x^{2}}{y^{2}} = \frac{x^{3}}{3} (\frac{2}{3} + \log x) + c$
D

$none of these$

Solution

$\begin{array}{l} We have x d y - y d x = x y^{3} (1 + \log x) d x \\ \Rightarrow - [\frac{y d x - x d y}{y^{2}}] = x y (1 + \log x) d x \\ Observing the complete differentials, we get - d (\frac{x}{y}) = x y (1 + \log x) d x \\ \Rightarrow \frac{- x}{y} d (\frac{x}{y}) = x^{2} (1 + \log x) d x \end{array}$
$\begin{array}{l} - {(\frac{x}{y})}^{2} = (1 + \log x) \frac{x^{3}}{3} - \int \frac{x^{3}}{3} \cdot \frac{1}{x} d x \\ \Rightarrow \frac{- x^{2}}{2 y^{2}} = \frac{x^{3}}{3} (1 + \log x) - \frac{x^{3}}{9} + \frac{c}{2} \\ \Rightarrow - \frac{x^{2}}{y^{2}} = \frac{2 x^{3}}{3} (\frac{2}{3} + \log x) + c \end{array}$

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