Question

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

Answer 1

$P(x,y)=bxe^{2xy};Q(x,y)=ye^{2xy}+x$

$P_x=be^{2xy}+2bye^{2xy}$

$Q_y=e^{2xy}+2xye^{2xy}$

The equation is exact if $P_x=Q_y$, hence

$be^{2xy}(1+2xy)=e^{2xy}(1+2xy)$

$b=1$

The solution of the equation $R(x,y)=const$ , where

$R=\int Pdy=\int xe^{2xy}dy=\frac{e^{2xy}}{2}+h(x)$

$R_x=Q$ hence

$ye^{2xy}+h'=ye^{2xy}+x$

$h'=x$

$h=\frac{x^2}{2}$

The solution of the equation

$\frac{1}{2}(e^{2xy}+x^2)=const$