Question

Test the DE(sin x . sin y – xey) dy = (ey + cos x.cos y)dx for exactness and solve it if exact

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\:(sinxsiny - x {e}^{y})dy = ({e}^{y} + cosxcosy)dx$

can be rewritten as

$\rm :\longmapsto\:({e}^{y} + cosxcosy)dx - (sinxsiny - x{e}^{y})dy = 0$

On Comparing with Mdx + Ndy = 0, we get

$\rm :\longmapsto\:M = {e}^{y} + cosxcosy$

and

$\rm :\longmapsto\:N = - sinxsiny + x{e}^{y}$

Now, Consider,

$\rm :\longmapsto\:M = {e}^{y} + cosxcosy$

So,

$\rm :\longmapsto\:\dfrac{\partial }{\partial y}M =\dfrac{\partial }{\partial y}( {e}^{y} + cosxcosy)$

$\rm :\longmapsto\:\dfrac{\partial M}{\partial y} =\dfrac{\partial }{\partial y}{e}^{y} +\dfrac{\partial }{\partial y} cosxcosy$

$\red{\rm :\longmapsto\:\dfrac{\partial M}{\partial y} = {e}^{y} - sinycosx - - (1)}$

Now, Consider,

$\rm :\longmapsto\:N = - sinxsiny + x{e}^{y}$

So,

$\rm :\longmapsto\:\dfrac{\partial }{\partial x}N =\dfrac{\partial }{\partial x}( - sinxsiny + x{e}^{y})$

$\rm :\longmapsto\:\dfrac{\partial N}{\partial x} = - \dfrac{\partial }{\partial x}sinxsiny +\dfrac{\partial }{\partial x} x{e}^{y}$

$\red{\rm :\longmapsto\:\dfrac{\partial N}{\partial x} = - cosxsiny + {e}^{y} - - - (2)}$

From equation (1), and equation (2), we concluded that

$\rm :\longmapsto\:\boxed{ \bf{ \:\dfrac{\partial N}{\partial x} = \dfrac{\partial M}{\partial y}}}$

So, given differential equation is exact.

So, Solution is

$\rm :\longmapsto\:\displaystyle\int_{y \: constant}M \: dx \: + \displaystyle\int_{term \: not \: containing \: x}N \: dy \: = \: c$

On substituting the values, we get

$\rm :\longmapsto\:\displaystyle\int_{y \: constant}({e}^{y} + sinxsiny) \: dx \: + \displaystyle\int_{term \: not \: containing \: x}0\: dy \: = \: c$

$\bf :\longmapsto\:x{e}^{y} - sinycosx = c$