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Question Number 40422 by rahul 19 last updated on 21/Jul/18

$$\mathrm{Solve}:$$$$\mathrm{yd}x-xdy+\log xdx=0$$

Commented by prof Abdo imad last updated on 22/Jul/18

$$\Rightarrow y-x\frac{dy}{dx}+ln\left(x\right)=0\Rightarrow {xy}^{{}^{\prime }}-y-ln\left(x\right)=0\Rightarrow$$$${xy}^{{}^{\prime }}-y=ln\left(x\right)$$$$he\Rightarrow {xy}^{{}^{\prime }}-y=0\Rightarrow \frac{y^{{}^{\prime }}}{y}=\frac{1}{x}\Rightarrow ln\mid y\mid =ln\mid x\mid +c\Rightarrow$$$$\Rightarrow y=k\mid x\mid$$$$case1x>0mvcmethodgive$$$$y^{{}^{\prime }}=k^{{}^{\prime }}x+k$$$$e)\Rightarrow x(k^{{}^{\prime }}x+k)-kx=ln\left(x\right)\Rightarrow$$$$k^{{}^{\prime }}{x}^2=ln\left(x\right)\Rightarrow k^{{}^{\prime }}=\frac{ln\left(x\right)}{x^2}\Rightarrow k\left(x\right)={\int }_{.}^x\frac{ln\left(t\right)}{t^2}dt+c$$$$byparts\int \frac{ln\left(t\right)}{t^2}dt=-\frac{1}{t}ln\left(t\right)+\int \frac{1}{t}\frac{dt}{t}$$$$=-\frac{1}{t}ln\left(t\right)-\frac{1}{t}\Rightarrow k\left(x\right)=-\frac{ln\left(x\right)}{x}-\frac{1}{x}+c\Rightarrow$$$$y\left(x\right)=x(-\frac{ln\left(x\right)}{x}-\frac{1}{x}+c)$$$$=cx-1-ln\left(x\right)$$$$case2x<0wefollowthesamemethod.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jul/18

$$\frac{xdy-ydx}{x^2}=\frac{lnxdx}{x^2}$$$$d\left(\frac{y}{x}\right)=\frac{lnxdx}{x^2}$$$$\int d\left(\frac{y}{x}\right)=lnx\int x^{-2}dx-\int \frac{1}{x}.\left(\frac{x^{-1}}{-1}\right)dx$$$$\frac{y}{x}=-\frac{lnx}{x}+\int \frac{dx}{x^2}+c$$$$\frac{y}{x}=-\frac{lnx}{x}-\frac{1}{x}+c$$$$\frac{y+1+lnx}{x}=c$$$$$$

Commented by rahul 19 last updated on 21/Jul/18

thank you sir ��

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