Question

Find the value of b for which the equation

(ye^2xy+x)dx +bxe^ 2xy dy= 0

Answer 1

Let us find the value of $b$ for which the equation $(ye^{2xy}+x)dx +bxe^{2xy} dy= 0$ is an exact differential equation. Since $\frac{\partial(ye^{2xy}+x)}{\partial y}=e^{2xy}+2xye^{2xy}$ and $\frac{\partial(bxe^{2xy})}{\partial x}=be^{2xy}+2bxye^{2xy}$, we conclude that $\frac{\partial(ye^{2xy}+x)}{\partial y}=\frac{\partial(bxe^{2xy})}{\partial x}$ implies $e^{2xy}+2xye^{2xy}=be^{2xy}+2bxye^{2xy}$, and hence $b=1.$

Answer: $b=1.$