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Sum

Solve : `(1+log x.y)dx +(1+x/y)`dy=0

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#### Solution

Compare given eqn with Mdx+Ndy=0

∴ M = (1+log x.y) `thereforeN=1+x/y`

`(delM)/(dely)=1/(xy)x=1/y` `delN)/(delx)=1/y`

`(delM)/(dely)=(delN)/(delx)`

Hence the given differential eqn is exact.

The solution of exact differential eqn is given by,

`intMdx+int(N-del/(dely)intMdx)dy=c` .....................(1)

`intMdx=int(1+logxy)dx=x+logxy.x-x=x.logxy`

`del/(dely)intMdx=x/y`

`int(N-del/delyintMdx)dy=int(1+x/y-x/y)dy=y`

From eqn (1), the solution of given differential eqn is ,

**x.log xy+y = c**

Concept: Exact Differential Equations

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