First slide

Methods of solving first order first degree differential equations

Question

The solution of $\frac{d y}{d x} = (x + y - 1) + \frac{x + y}{\log (x + y)}$ is given by

Moderate

A

$\{1 + \log (x + y)\} - \log \{1 + \log (x + y)\} = x + c$
B

$\{1 + \log (x + y)\} - \log \{1 - \log (x + y)\} = x + c$
C

${\{1 + \log (x + y)\}}^{2} - \log \{1 + \log (x + y)\} = x + c$
D

none of these

Solution

$\begin{array}{l} \frac{d y}{d x} = (x + y - 1) + \frac{x + y}{\ln (x + y)} \\ Let x + y = z \\ \Rightarrow 1 + \frac{d y}{d x} = \frac{d z}{d x} \Rightarrow \frac{d z}{d x} - 1 = z - 1 + \frac{z}{\log z} \\ \Rightarrow \frac{d z}{d x} = \frac{z}{\log z} + z = z (\frac{1 + \log z}{\log z}) \\ \Rightarrow \int \frac{\log z d z}{z (1 + \log z)} = \int d x \\ Let \log z = t \\ \Rightarrow \int \frac{t}{1 + t} d t = \int d x \\ \Rightarrow t - \ln (1 + t) = x + c_{0} \\ \Rightarrow \log (x + y) - \log (1 + \log (x + y)) = x + C_{0} \end{array}$
Regarding the expression by adding 1 both side, we get
$1 + l o g (x + y) - l o g (1 + l o g (x + y)) = x + c, w h e r e c = 1 + C_{0}$

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