Question

(2x log x-xy)dy+2ydx=0
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Answer 1

(2x log x-xy)dy+2ydx=0

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M = 2y

N = 2x log x - xy

$\dfrac{\partial M}{\partial y}=2$

$\dfrac{\partial N}{\partial y} = 2(1+ log x)-y$

Now, $\dfrac {\frac{\partial M}{\partial y} - \frac{\partial N}{\partial y}}{N}$

$= \dfrac { 2 - [2(1+logx)-y]} {2x log x - xy}\\\\\\= \dfrac{-2log x +y}{2xlog x - xy}\\\\\\\dfrac{-1}{x} = f(x)$

$I.F = e^{\int f(x) dx}\\\\= e^{\int \frac {1}{x} dx}\\\\=e^{-log x}\\\\=e^{log x^{-1}}\\\\=x^{-1}\\\\=\dfrac {1}{x}$

Upon multiplying the given equation with 1/x, we get,

$= \dfrac{1}{x} 2y dx + \dfrac{1}{x} (2log x - y) dy=0$

$\int \dfrac{2y}{x} dx + \int -y dy = c\\\\2ylog x - \dfrac{1}{2}y^2 =c$

Hence the result.

Answer 2

Answer:

(2x log x-xy)dy+2ydx=0

M = 2y

N = 2x log x - xy

\dfrac{\partial M}{\partial y}=2∂y∂M=2

\dfrac{\partial N}{\partial y} = 2(1+ log x)-y∂y∂N=2(1+logx)−y

Now, \dfrac {\frac{\partial M}{\partial y} - \frac{\partial N}{\partial y}}{N}N∂y∂M−∂y∂N

\begin{gathered}= \dfrac { 2 - [2(1+logx)-y]} {2x log x - xy}\\\\\\= \dfrac{-2log x +y}{2xlog x - xy}\\\\\\\dfrac{-1}{x} = f(x)\end{gathered}=2xlogx−xy2−[2(1+logx)−y]=2xlogx−xy−2logx+yx−1=f(x)

\begin{gathered}I.F = e^{\int f(x) dx}\\\\= e^{\int \frac {1}{x} dx}\\\\=e^{-log x}\\\\=e^{log x^{-1}}\\\\=x^{-1}\\\\=\dfrac {1}{x}\end{gathered}I.F=e∫f(x)dx=e∫x1dx=e−logx=elogx−1=x−1=x1

Upon multiplying the given equation with 1/x, we get,

= \dfrac{1}{x} 2y dx + \dfrac{1}{x} (2log x - y) dy=0=x12ydx+x1(2logx−y)dy=0

\begin{gathered}\int \dfrac{2y}{x} dx + \int -y dy = c\\\\2ylog x - \dfrac{1}{2}y^2 =c\end{gathered}∫x2ydx+∫−ydy=c2ylogx−21y2=c

Hence the result.