First slide

Methods of solving first order first degree differential equations

Question

If $x^{2} y d x - (x^{3} + y^{3}) d y = 0$ , $y (0) = 1$ and $y^{3} \log y = k x^{3}$ then k =

Moderate

A

$\frac{1}{2}$
B

$\frac{1}{3}$
C

1
D

2

Solution

$\begin{array}{l} \Rightarrow \frac{d y}{d x} = \frac{x^{2} y}{x^{3} + y^{3}} = \frac{y / x}{1 + (y / x)^{3}} \\ V + x \frac{d v}{d x} = \frac{v}{1 + v^{3}} (put y = v x) \\ \Rightarrow x \frac{d v}{d x} = \frac{- v^{4}}{1 + v^{3}} \\ \int \frac{1 + v^{3}}{v^{4}} d v = - \int \frac{d x}{x} \\ \Rightarrow \int (v^{- 4} + \frac{1}{v}) d v = - \int \frac{1}{x} d x \\ \Rightarrow \frac{v^{- 3}}{- 3} + \ln v = - \ln x + c \end{array}$
$\begin{array}{l} \Rightarrow \ln |x (\frac{y}{x})| - \frac{1}{3 v^{3}} = c \\ \ln y - \frac{x^{3}}{3 y^{3}} = c \\ y (0) = 1 \Rightarrow c = 0 \end{array}$
$∴ \log y = \frac{x^{3}}{3 y^{3}}$

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