Question

Solve $\displaystyle\rm \: xy \frac{dy}{dx} = y + 2$ given that x=2 when y=1.
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Answer 1

$\large \mathbb{ \green{GIVEN : - }}$

$\displaystyle\rm \: xy \frac{dy}{dx} = y + 2$

$\displaystyle\rm \: And \: \: x = 2 \: \: \: \: y = 1$

$\large \mathbb{ \green{SOLUTION : - }}$

$\displaystyle\rm \rightarrow \: xy \frac{dy}{dx} = y + 2$

$\displaystyle\rm \rightarrow \: \frac{y.dy}{(y + 2)} = \frac{dx}{x}$

$\displaystyle\rm \rightarrow \: \frac{(y + 2 - 2).dy}{(y + 2)} = \frac{dx}{x}$

$\displaystyle\rm \rightarrow \: \frac{(y + 2)dy}{(y + 2)} - \frac{2dy}{(y + 2)} = \frac{dx}{x}$

$\displaystyle\rm \rightarrow \: dy - \frac{2.dy}{(y + 2)} = \frac{dx}{x}$

$\rm \: On \: integrating \: both \: side$

$\displaystyle\rm \rightarrow \: \int \bigg(dy - \frac{2.dy}{(y + 2)} \bigg) = \int \frac{dx}{x}$

$\displaystyle\rm \rightarrow \: y - 2 \: log(y + 2) = log(x) + log(c)$

$\displaystyle \pink{\rm \: at \: \: \: x = 2 \: \: \: \: y = 1}$

$\displaystyle\rm \rightarrow \: 1 - 2 \: log(1 + 2) = log(2) + log(c)$

$\displaystyle\rm \rightarrow \: 1 - 2 \: log(3) = log(2) + log(c)$

$\displaystyle\rm \rightarrow \: 1 - \: log( {3}^{2} ) - log(2) = log(c)$

$\displaystyle\rm \rightarrow \: log(e) - \: log( 9 ) - log(2) = log(c)$

$\displaystyle\rm \rightarrow \: log \bigg( \frac{e}{2 \times 9} \bigg) = log(c)$

$\displaystyle\rm \rightarrow \: log(c) = log(\frac{e}{18} )$

$\displaystyle\rm \rightarrow \: y - 2 \: log(y + 2) = log(x) + log( \frac{e}{18} )$

$\displaystyle \boxed{\rm \rightarrow \: y = log \bigg( \frac{ex {(y + 2)}^{2}}{18} \bigg) }$