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Methods of solving first order first degree differential equations

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Question

$If x \frac{d y}{d x} = x^{2} + y - 2, y (1) = 1 then y (2) is$

Moderate

Solution

$\frac{d y}{d x} = x + \frac{y}{x} - \frac{2}{x}$
$\begin{array}{l} \frac{dy}{dx} - \frac{1}{x} y = (x - \frac{2}{x}), I.F. = e^{- \int \frac{1}{x} dx} = e^{- \ln x} = x^{- 1} = \frac{1}{x} \\ GS of the D.E is \frac{y}{x} = \int (x - \frac{2}{x}) \frac{1}{x} dx \\ \frac{y}{x} = \int 1 d x - 2 \int \frac{1}{x^{2}} d x \\ \frac{y}{x} = x + \frac{2}{x} + c \\ Put x=1, y=1 \\ 1 = 3 + c \to c = - 2 \\ \frac{y}{x} = \frac{x^{2} + 2 - 2 x}{x} \\ \to y = x^{2} - 2 x + 2 \\ \to y (2) = 2^{2} - 4 + 2 = 2 \end{array}$

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