I have it down toln⁡(y)+(xy)y′=y′ln⁡(x)+yxThe problem is factoring out y', which leads to eithery′=ln⁡(y)−(yx)ln⁡(x)−(xy)or toy′=(yx)−ln⁡(y)(xy)−ln⁡(x)Am I missing something?

xy=yx⇒ylog⁡x=xlog⁡y⇒log⁡xdy+yxdx=log⁡ydx+xydy⇒(log⁡x−xy)dy=(log⁡y−yx)dx⇒dydx=log⁡y−yxlog⁡x−xy

Writeeyln⁡x=exln⁡yThen(yx+y′ln⁡x)eyln⁡x=(ln⁡y+xy′y)exln⁡yrearranging, we get(xyln⁡x−xyyx)y′=yxln⁡y−yxxyUse yx=xy and gety′=ln⁡y−(yx)ln⁡x−(xy)

xy=yx⇒ylog⁡x=xlog⁡y⇒log⁡xdy+yxdx=log⁡ydx+xydy⇒(log⁡x−xy)dy=(log⁡y−yx)dx⇒dydx=log⁡y−yxlog⁡x−xy

Writeeyln⁡x=exln⁡yThen(yx+y′ln⁡x)eyln⁡x=(ln⁡y+xy′y)exln⁡yrearranging, we get(xyln⁡x−xyyx)y′=yxln⁡y−yxxyUse yx=xy and gety′=ln⁡y−(yx)ln⁡x−(xy)

Best solutions to - "I have it down toln⁡(y)+(xy)y′=y′ln⁡(x)+yxThe problem is factoring..."

Jamya Shea

2022-08-20

I have it down to

\ln (y) + (\frac{x}{y}) y^{'} = y^{'} \ln (x) + \frac{y}{x}

The problem is factoring out y', which leads to either

y^{'} = \frac{\ln (y) - (\frac{y}{x})}{\ln (x) - (\frac{x}{y})}

or to

y^{'} = \frac{(\frac{y}{x}) - \ln (y)}{(\frac{x}{y}) - \ln (x)}

Am I missing something?

Terry Avery

Beginner2022-08-21Added 10 answers

x^{y} = y^{x} \Rightarrow y \log x = x \log y \Rightarrow \log x d y + \frac{y}{x} d x = \log y d x + \frac{x}{y} d y \Rightarrow

(\log x - \frac{x}{y}) d y = (\log y - \frac{y}{x}) d x \Rightarrow \frac{d y}{d x} = \frac{\log y - \frac{y}{x}}{\log x - \frac{x}{y}}

decorraj9

Beginner2022-08-22Added 1 answers

Write

e^{y \ln x} = e^{x \ln y}

Then

(\frac{y}{x} + y^{'} \ln x) e^{y \ln x} = (\ln y + \frac{x y^{'}}{y}) e^{x \ln y}

rearranging, we get

(x^{y} \ln x - \frac{x}{y} y^{x}) y^{'} = y^{x} \ln y - \frac{y}{x} x^{y}

Use

y^{x} = x^{y}

and get

y^{'} = \frac{\ln y - (\frac{y}{x})}{\ln x - (\frac{x}{y})}

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