First slide

Methods of solving first order first degree differential equations

Question

$Solution of differential equation \frac{d t}{d x} = \frac{t (\frac{d}{d x} (g (x))) - t^{2}}{g (x)} is$

Moderate

A

$t = \frac{g (x)}{x} + c$
B

$t = \frac{g (x)}{x^{2}} + c$
C

$t = \frac{g (x)}{x + c}$
D

$t = g (x) + x + c$

Solution

$\begin{array}{l} Rearranging the terms of equation, \\ We get \frac{d t}{d x} - t \frac{g^{'} (x)}{g (x)} = \frac{t^{2}}{g (x)} \\ \Rightarrow \frac{1}{t^{2}} \frac{d t}{d x} + \frac{1}{t} \frac{g^{'} (x)}{g (x)} = \frac{1}{g (x)} \dots \dots \dots \dots . . (1) \\ Let z = \frac{1}{t} \Rightarrow - \frac{1}{t^{2}} \frac{d t}{d x} = \frac{d z}{d x} \end{array}$
$\begin{array}{l} Thus, from (i) we obtain \frac{d z}{d x} + \frac{g^{'} (x)}{g (x)} z = \frac{1}{g (x)} \\ Which if clearly linear in 2 and \frac{d z}{d x} with \\ I . F = e^{\int \frac{g^{'} (x)}{g (x)} d x} = e^{\log [g (x)]} = g (x) \end{array}$
$\Rightarrow Thus complete solution is z. g (x) = \int g (x) \cdot \frac{1}{g (x)} d x + c$
$\Rightarrow \frac{1}{t} g (x) = x + c \Rightarrow \frac{g (x)}{x + c} = t$

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