First slide

Methods of solving first order first degree differential equations

Question

$A curve passes through (2,0) and slope of the tangent at the point p (x, y) i s \frac{(x + 1)^{2} + y - 3}{x + 1}, the area bounds by$
$the curve and the x -axis is$

Moderate

A

1
B

2
C

2/3
D

4/3

Solution

$\begin{array}{l} \frac{dy}{dx} = x + 1 + \frac{y - 3}{x + 1}, put X = x + 1, Y = y - 3 \\ \Rightarrow \frac{d Y}{d X} - \frac{Y}{X} = X \end{array}$
$\begin{array}{l} I . F = e^{- \int \frac{1}{X} d x} = e^{- \log X} = \frac{1}{X} \\ Solution of the D.E in \frac{Y}{X} = \int X \frac{1}{X} d x \\ \frac{Y}{X} = X + c \\ \Rightarrow \frac{y - 3}{x + 1} = x + 1 + c - - - - (1) \\ If x = 2, y = 0 \Rightarrow \frac{- 3}{3} = 3 + c \Rightarrow c = - 4 \\ s u b i n (1) \\ ∴ y - 3 = (x + 1)^{2} - 4 (x + 1) = x^{2} - 2 x - 3 \\ \Rightarrow y = x^{2} - 2 x meets x -axis ⇒y=0 x = 0, 2 \\ Hence the required area = \int_{0}^{2} (2 x - x^{2}) dx = 4 - \frac{8}{3} = \frac{4}{3} \end{array}$

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