Question

Find the value of b for which the equation
( ye^(2xy)+x)dx+bxe^(2xy)dy= 0
is exact, and hence solve it for that value of b

Answer 1

The given equation is $(ye^{2xy}+x)dx+bxe^{2xy}dy=0$ .

We know that a differential equation $Mdx+Ndy=0$ is exact if

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

Therefore, according to question,

$M=(ye^{2xy}+x) \ and \ N=bxe^{2xy}$

Now,

$$\frac{\partial (ye^{2xy}+x)}{\partial y}=e^{2xy}+2xye^{2xy}=(1+2xy)e^{2xy}.......(1)$$

and

$$\frac{\partial (bxe^{2xy})}{\partial x}=be^{2xy}+2xybe^{2xy}=(1+2xy)be^{2xy}........(2)$$

from equation (1) and (2) ,we have the given differential equation is exact if $b=1$ .

Now , putting b=1 in the given differential equation we get

$(ye^{2xy}+x)dx+xe^{2xy}dy=0$

$\implies \ ye^{2xy}dx+xdx+xe^{2xy}dy=0$

$\implies \ e^{2xy}(ydx+xdy)+xdx=0$

$\implies \ e^{2xy}d(xy)+xdx=0$

Integrating, we get

$\frac{1}{2}e^{2xy}+\frac{x^2}{2}=c$

where "c" is an integration constant.