First slide

Methods of solving first order first degree differential equations

Question

The solution of the differential equation $\frac{d y}{d x} = \frac{{(x + y)}^{2}}{(x + 2) (y - 2)}$ is

Moderate

A

$e^{\frac{2 (y - 2)}{x + 2}} = = c^{2} (x + 2)^{3} (x + 2 y - 2)$ .
B

$e^{\frac{2 (y - 2)}{x + 2}} = c^{2} (x + 2)^{4} (x + 2 y - 2)$ ..
C

$e^{\frac{2 (y - 2)}{x + 2}} = c^{2} (x + 2)^{3} (x + 2 y + 2)$ …
D

$e^{\frac{2 (y - 2)}{x + 2}} = c^{2} (x + 2)^{3} (x - 2 y + 2)$

Solution

$\begin{array}{l} \frac{d y}{d x} = \frac{(x + y)^{2}}{(x + 2) (y - 2)} = \frac{[(x + 2) + (y - 2)]^{2}}{(x + 2) (y - 2)} \\ put X = x + 2 and Y = y + 2 \\ \frac{d Y}{d X} = \frac{(X + Y)^{2}}{X Y} \\ p u t Y = v X \\ v + X \frac{d v}{d X} = \frac{v^{2} + 2 v + 1}{v} = v + \frac{2 v + 1}{v} \\ \Rightarrow X \frac{d v}{d x} = \frac{2 v + 1}{v} \\ \Rightarrow \frac{v}{2 v + 1} d v = \frac{d x}{X} \\ \Rightarrow \int (1 - \frac{1}{1 + 2 v}) d v = 2 \int \frac{d x}{X} \end{array}$
$\begin{array}{l} v - \frac{1}{2} \ln | 1 + 2 v | = 2 \ln X + \ln c \\ 2 v - \ln |1 + 2 v| = 4 \ln X + 2 \ln c \\ 2 v = \ln |c^{2} X^{4} (1 + 2 v)| \\ \frac{2 Y}{X} = \ln |c^{2} X^{4} (1 + \frac{2 Y}{X})| \\ e^{\frac{2 Y}{X}} = |c^{2} X^{3} (X + 2 Y)| \\ \Rightarrow e^{\frac{2 (y - 2)}{x + 2}} = c^{2} (x + 2)^{3} (x + 2 + 2 y - 4) = c^{2} (x + 2)^{3} (x + 2 y - 2) \end{array}$

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