Question

xlogx dy/dx+y=logx2​

Answer 1

Answer:

thank you for your free points...xD

Answer 2

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Given:

xlogx dy/dx +y=logx²

Formula:

$\bf{ \bullet \ \frac{dy}{dx} = x = 1}$

$\: \bf \bullet \: \frac{dy}{dx} = logx = \frac{1}{x}$

$\: \bf{ \bullet \: \frac{dy}{dx} = log {x}^{2} = 2logx \times \frac{1}{x} } \: \: \: \:$

$\: \bf{ \bullet \: \frac{dy}{dx} = y = dy}$

•Apply the formula in this equation :

$\: \implies \displaystyle \sf \: xlogx+ \frac{dy}{dx} + y = log {x}^{2}$

$\: \implies \displaystyle \sf \: xlogx + y - log {x}^{2} = \frac{dy}{dx}$

Differentiating the equation with respect to x:

$\: \implies \displaystyle \sf \: xlogx + y - log {x}^{2} = \frac{dy}{dx}$

$\: \\ \implies \displaystyle \sf \: x \frac{1}{x} + \frac{dy}{dx} - 2logx \times \frac{1}{x} = \frac{dy}{dx}$

$\: \: \implies \displaystyle \sf \: 1 + \frac{dy}{dx} - \frac{2logx}{x} = \frac{dy}{dx}$