First slide

Types of solutions of differential equations

Question

$The curve satisfying the equation \frac{d y}{d x} = \frac{y (x + y^{3})}{x (y^{3} - x)} and passing through the point (4, - 2) is$

Moderate

A

$y^{2} = 2 x$
B

$y = - x$
C

$y^{3} = - 2 x$
D

$None of these$

Solution

$\begin{array}{l} Given differential equation is \\ \begin{array}{r} \begin{array}{l} \frac{d y}{d x} = \frac{y (x + y^{3})}{x (y^{3} - x)} \\ x y^{3} d y - y^{4} d x = x y d x + x^{2} d y \\ x y^{3} d y - y^{4} d x = x (y d x + x d y) \\ \frac{x y^{3} d y - y^{4} d x}{x} = (y d x + x d y) \end{array} \end{array} \end{array}$
$\begin{array}{l} \Rightarrow x (\frac{x y^{3} d y - y^{4} d x}{x^{2}}) = d (x y) \\ \Rightarrow x y^{3} (\frac{x d y - y d x}{x^{2}}) = d (x y) \\ \Rightarrow x y^{3} [d (\frac{y}{x})] = d (x y) \\ (x^{2} y^{2}) \cdot (\frac{y}{x}) d (\frac{y}{x}) = d (x y) \end{array}$
$\begin{array}{l} \Rightarrow (\frac{y}{x}) d (\frac{y}{x}) = \frac{1}{(x y)^{2}} d (x y) \\ \Rightarrow \frac{1}{2} d {(\frac{y}{x})}^{2} = d [(x y)^{- 1}] \\ \Rightarrow \frac{1}{2} {(\frac{y}{x})}^{2} = - \frac{1}{x y} + c \\ As it passes through (4, - 2) \end{array}$
$\begin{aligned} \Rightarrow \frac{1}{2} {(\frac{- 2}{4})}^{2} = \frac{- 1}{4 (- 2)} + c \Rightarrow \frac{1}{2} (\frac{1}{4}) = \frac{1}{8} + c \Rightarrow c = 0 \\ \frac{1}{2} (\frac{y^{2}}{x^{2}}) = \frac{1}{x y} \Rightarrow y^{2} = \frac{- 2 x^{2}}{x y} \Rightarrow y^{2} = \frac{- 2 x}{y} \Rightarrow y^{3} = - 2 x \end{aligned}$

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