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Methods of solving first order first degree differential equations

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Question

$The solution of the differential equation (2 x - 4 y + 3) \frac{d y}{d x} + (x - 2 y + 1) = 0 is$ $\log [4 (x - 2 y) + 5] = λ (x - 2 y) + c then λ is$

Moderate

Solution

$\begin{array}{l} \frac{d y}{d x} = \frac{- (x - 2 y + 1)}{2 (x - 2 y) + 3} \\ Put x - 2 y = t \\ 1 - 2 \frac{d y}{d x} = \frac{d t}{d x} \\ \Rightarrow \frac{1}{2} (1 - \frac{d t}{d x}) = \frac{d y}{d x} \\ \Rightarrow \frac{1}{2} (1 - \frac{d t}{d x}) = - \frac{(t + 1)}{2 t + 3} \\ \Rightarrow 1 - \frac{d t}{d x} = \frac{- 2 t - 2}{2 t + 3} \\ \frac{d t}{d x} = 1 + \frac{2 t - 2}{2 t + 3} = \frac{4 t + 5}{2 t + 3} \\ \Rightarrow \frac{1}{2} \int (1 + \frac{1}{4 t + 5}) d t = \int d x \end{array}$
$\Rightarrow \frac{1}{4} \log | 4 (x - 2 y) + 5 | = x + 2 y + c / 4$
$\begin{array}{l} \Rightarrow \log | 4 (x - 2 y) + 5 | = 4 (x + 2 y) + c \\ \Rightarrow λ = 4 \end{array}$

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