First slide

Methods of solving first order first degree differential equations

Question

$A solution curve of the differential equation (x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0, passes$ through the point $(1, 3)$ . Then the solution curve is

Difficult

A

$intersects y = x + 2 exactly at one point$
B

$intersects y = x + 2 exactly at two points$
C

$intersects y = (x + 2)^{2}$
D

$does NOT intersect y = (x + 3)^{2}$

Solution

$G i v e n e q u a t i o n i s [{(x + 2)}^{2} + y (x + 2)] \frac{d y}{d x} - y^{2} = 0$
$L e t x + 2 = X, y = Y$
$\begin{array}{l} (X^{2} + X Y) d Y - Y^{2} d X = 0 \\ \frac{d Y}{d X} = \frac{Y^{2}}{X^{2} + X Y} \\ Y = v X \Rightarrow \frac{d v}{d X} + v = \frac{v^{2}}{v + 1} \\ \Rightarrow X \frac{d v}{d X} = \frac{- v}{v + 1} \\ \Rightarrow \frac{v + 1}{v} d v = \frac{- 1}{X} d X \\ \Rightarrow v + \log v = - \log X + \log c \\ \Rightarrow \frac{Y}{X} + \log Y = \log c \\ ⇒ \frac{y}{x + 2} + \log y = \log c \\ i t p a s s e s t h r o u g h (1, 3) \Rightarrow 1 + \log 3 = \log c \\ t h e r e f o r e \frac{y}{x + 2} + \log y = 1 + \log 3 \\ o p t i o n (1) : i f y = x + 2 t h e n \frac{x + 2}{x + 2} + \log y = 1 + \log 3 \\ \log y = \log 3 \Rightarrow y = 3 h e n c e (1, 3) i s c o m m o n s o l u t i o n \\ o p t i o n (3) : i f y = {(x + 2)}^{2} t h e n \frac{{(x + 2)}^{2}}{x + 2} + \log {(x + 2)}^{2} = 1 + \log 3 \\ \Rightarrow x + 2 + \log {(x + 2)}^{2} = 1 + \log 3, n o v a l u e o f x s a t i s f i e s . \\ o p t i o n (4) : i f y = {(x + 3)}^{2} t h e n \frac{{(x + 3)}^{2}}{x + 2} + \log {(x + 3)}^{2} = 1 + \log 3, n o v a l u e o f x s a t i s f i e s . \end{array}$

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