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General Solution of a Differential Equation

If xdydx=x2+y...

Question

If $x \frac{d y}{d x} = x^{2} + y - 2, y (1) = 1,$ and $x > 0$ then $y (2)$ equals

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Solution

Given $\frac{d y}{d x} - \frac{1}{x} y = (x - \frac{2}{x})$
I.F. $= e^{- \int \frac{1}{x} d x} = e^{- ln | x |} = \frac{1}{x}$
Now general solution is given by $\frac{y}{x} = \int (x - \frac{2}{x}) \frac{1}{x} d x$
$\Rightarrow \frac{y}{x} = x + \frac{2}{x} + C$
As $y (1) = 1 \Rightarrow C = - 2$
$∴ \frac{y}{x} = x + \frac{2}{x} - 2 \Rightarrow y = x^{2} - 2 x + 2$
Hence, $y (2) = {(2)}^{2} - 2 (2) + 2 = 2$

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General Solution of a Differential Equation

Standard XII Mathematics

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