Solve : ( 1 + Log X . Y ) D X + ( 1 + X Y ) Dy=0 - Applied Mathematics 2

Sum

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

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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