Question

Solve : ${\displaystyle (1+\log x.y)dx+(1+\frac{x}{y})}$dy=0

Answer 1

Compare given eqn with Mdx+Ndy=0

∴ M = (1+log x.y)               ${\displaystyle \therefore N=1+\frac{x}{y}}$

${\displaystyle \frac{\partial M}{\partial y}=\frac{1}{xy}x=\frac{1}{y}}$    ${\displaystyle \partial N\frac{)}{\partial x}=\frac{1}{y}}$

${\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}}$

Hence the given differential eqn is exact.
The solution of exact differential eqn is given by,

${\displaystyle \int Mdx+\int (N-\frac{\partial }{\partial y}\int Mdx)dy=c}$      .....................(1)

${\displaystyle \int Mdx=\int (1+\log xy)dx=x+\log xy.x-x=x.\log xy}$

${\displaystyle \frac{\partial }{\partial y}\int Mdx=\frac{x}{y}}$

${\displaystyle \int (N-\frac{\partial }{\partial }y\int Mdx)dy=\int (1+\frac{x}{y}-\frac{x}{y})dy=y}$

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

x.log xy+y = c