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Question Number 95098 by i jagooll last updated on 23/May/20

$$[\frac{\mathrm{y}}{{\mathrm{x}}^2+{\mathrm{y}}^2}+\frac{\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{y}}^2}]\mathrm{dx}+[\frac{\mathrm{y}}{{\mathrm{x}}^2+{\mathrm{y}}^2}-\frac{\mathrm{x}}{{\mathrm{x}}^2+{\mathrm{y}}^2}]\mathrm{dy}=0$$

Answered by bobhans last updated on 23/May/20

$$\mathrm{let}\mathrm{x}=\mathrm{r}\cos \theta ;\mathrm{y}=\mathrm{r}\sin \theta ;{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$$$$\mathrm{dx}=\cos \theta \mathrm{dr}-\mathrm{rsin}\theta \mathrm{d}\theta ;\mathrm{dy}=\mathrm{rcos}\theta \mathrm{d}\theta +\sin \theta \mathrm{dr}$$$$\Rightarrow (\frac{\sin \theta }{\mathrm{r}}+\frac{\cos \theta }{\mathrm{r}})(\cos \theta \mathrm{dr}-\mathrm{rsin}\theta \mathrm{d}\theta )+$$$$(\frac{\sin \theta }{\mathrm{r}}-\frac{\cos \theta }{\mathrm{r}})(\sin \theta \mathrm{dr}+\mathrm{rcos}\theta \mathrm{d}\theta )=0$$$$\Rightarrow (\sin \theta +\cos \theta )(\mathrm{rcos}\theta \mathrm{dr}-{\mathrm{r}}^2\sin \theta \mathrm{d}\theta )+$$$$(\sin \theta -\cos \theta )(\mathrm{rsin}\theta \mathrm{dr}+{\mathrm{r}}^2\cos \theta \mathrm{d}\theta )=0$$$$\Rightarrow \frac{1}{2}\mathrm{rsin}2\theta \mathrm{dr}-{\mathrm{r}}^2\sin {}^2\theta \mathrm{d}\theta +\mathrm{rcos}{}^2\theta \mathrm{dr}-\frac{1}{2}{\mathrm{r}}^2\sin 2\theta \mathrm{d}\theta$$$$+\mathrm{r}\sin {}^2\theta \mathrm{dr}+\frac{1}{2}{\mathrm{r}}^2\sin 2\theta \mathrm{d}\theta -\frac{1}{2}\mathrm{rsin}2\theta \mathrm{dr}-{\mathrm{r}}^2\cos {}^2\theta \mathrm{d}\theta =0$$$$\Rightarrow \mathrm{r}\cos {}^2\theta \mathrm{dr}+\mathrm{rsin}{}^2\theta \mathrm{dr}={\mathrm{r}}^2(\sin {}^2\theta +\cos {}^2\theta )\mathrm{d}\theta$$$$\Rightarrow \mathrm{dr}=\mathrm{r}\mathrm{d}\theta ;\int \frac{\mathrm{dr}}{\mathrm{r}}=\theta +\mathrm{c}$$$$\ln \mathrm{r}=\theta +\mathrm{c}\Rightarrow \mathrm{r}={\mathrm{Ce}}^{\theta }\Rightarrow \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}=\pm {\mathrm{Ce}}^{{\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}.$$$$\mathrm{done}!!$$$$$$

Commented by i jagooll last updated on 23/May/20

$$\mathrm{waw}..=\mathrm{great}$$$$$$

Commented by peter frank last updated on 23/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by peter frank last updated on 23/May/20

$$\mathrm{thank}\mathrm{you}$$

Answered by mr W last updated on 23/May/20

$$(y+x)dx+(y-x)dy=0$$$$\frac{dy}{dx}=\frac{x+y}{x-y}$$$$lety=xu$$$$u+x\frac{du}{dx}=\frac{1+u}{1-u}$$$$x\frac{du}{dx}=\frac{1+u^2}{1-u}$$$$\int \frac{1-u}{1+u^2}du=\int \frac{dx}{x}$$$${\tan }^{-1}u-\frac{1}{2}\ln (1+u^2)=\ln x+C$$$$\Rightarrow 2{\tan }^{-1}\frac{y}{x}=\ln (x^2+y^2)+C$$

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