Question

True or False: Every separable first-order equation $\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{h}\mathbf{\left(}\mathit{y}\mathbf{\right)}$ is exact.

Short answer

The given statement is true.

Step-by-step solution

The first-order equation

Let's consider the separable equation:

$\frac{dy}{dx}=g\left(x\right)h\left(y\right)\Rightarrow -g\left(x\right)dx+\frac{dy}{h\left(y\right)}=0$

And non-linear first-order DE:

$M(x,y)dx+N(x,y)dy=0$

To determine the differential equation

To determine whether a given differential equation is exact, we use test for Exactness in which states:

DE is exact, which implies that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

Comparing separable equation with non-linear first-order DE, we get the following:

$M(x,y)=-g\left(x\right),N(x,y)=\frac{1}{h\left(y\right)}$

Now, since

$\frac{\partial M}{\partial y}=0,\frac{\partial N}{\partial x}=0$

Final proof

We get, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

Therefore, we can conclude that every separable first-order equation $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is exact.

The given statement is true.