u-R-2-R-u-y-u-x-x-2-y-2-2xu-0-

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Question Number 304 by 123456 last updated on 20/Dec/14

$$u:{\mathbb{R}}^2\to \mathbb{R}$$$$\frac{\partial u}{\partial y}+\frac{\partial u}{\partial x}(x^2+y^2)+2xu=0$$

Commented by prakash jain last updated on 20/Dec/14

$$\mathrm{Let}\mathrm{us}\mathrm{consider}\mathrm{this}\mathrm{equation}$$$$dx-(x^2+y^2)dy=0\left(\mathrm{i}\right)$$$$\mathrm{This}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{exact}\mathrm{equation}.$$$$\mathrm{M}=1,N=-(x^2+y^2)$$$$\mathrm{This}\mathrm{equation}\mathrm{can}\mathrm{be}\mathrm{converted}\mathrm{to}\mathrm{exact},\mathrm{if}$$$$\frac{\partial M}{\partial y}.u+\frac{\partial u}{\partial y}\cdot M=\frac{\partial N}{\partial x}\cdot u+\frac{\partial u}{\partial x}N$$$$\frac{\partial u}{\partial y}=-2xu-(x^2+y^2)\frac{\partial u}{\partial x}$$$$\mathrm{Which}\mathrm{is}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}.$$$$\mathrm{In}\mathrm{other}\mathrm{words}\mathrm{the}\mathrm{given}\mathrm{differential}$$$$\mathrm{equation}\mathrm{gives}\mathrm{integrating}\mathrm{factor}\mathrm{for}\left(\mathrm{i}\right).$$$$\frac{\partial M}{\partial y}=0,\frac{\partial N}{\partial x}=-2x$$$$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\frac{-2x}{-(x^2+y^2)}\mathrm{This}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{function}\mathrm{of}\mathrm{only}x.$$$$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M}=\frac{-2x}{1}\mathrm{This}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{function}\mathrm{of}\mathrm{only}y.$$$$\mathrm{So}\mathrm{it}\mathrm{does}\mathrm{not}\mathrm{look}\mathrm{like}u(x,y)\mathrm{is}\mathrm{an}\mathrm{elementry}$$$$\mathrm{function}.$$$$\mathrm{I}\mathrm{will}\mathrm{see}\mathrm{if}\mathrm{it}\mathrm{can}\mathrm{he}\mathrm{solved}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{other}$$$$\mathrm{functions}.$$

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