First recall that if $z = f(x, y)$ is a differentiable function of $x$ and $y$, where $x = g(t)$ and $y = h(t)$ are both differentiable functions of $t$, then $z$ is a differentiable function of $t$ whose derivative is given by the chain rule:

Now suppose the equation \[ f(x,y)= C \] defines $y$ implicitly as a function of $x$ (here $C$ is a constant). Then $y = y(x)$ can be shown to satisfy a first order ODE obtained by using the chain rule above.

Now suppose the equation \[ f(x,y)= C \] defines $y$ implicitly as a function of $x$ (here $C$ is a constant). Then $y = y(x)$ can be shown to satisfy a first order ODE obtained by using the chain rule above. In this case, $z = f(x, y(x)) = C,$ so

A first order ODE of the form \[ P\left(x,y\right)+ Q\left(x,y\right) \frac{dy}{dx}=0 \qquad (8) \] is called exact, if there is a function $\,f(x, y)$ (compare $(8)$ with $(7)$ above) such that \[ f_x\left(x,y\right) = P\left(x,y\right) \quad \text{and}\quad f_y\left(x,y\right) = Q\left(x,y\right). \]

The solution is then given implicitly by the equation $\,f\left(x,y\right)=C.$ The constant $C$ can usually be determined by some kind of "initial condition".

A first order ODE of the form \[ P\left(x,y\right)+ Q\left(x,y\right)\frac{dy}{dx}=0 \qquad (8) \] is called exact, if there is a function $\,f(x, y)$ (compare $(8)$ with $(7)$ above) such that \[ f_x\left(x,y\right) = P\left(x,y\right) \quad \text{and}\quad f_y\left(x,y\right) = Q\left(x,y\right). \]

The solution is then given implicitly by the equation $\,f\left(x,y\right)=C.$ The constant $C$ can usually be determined by some kind of “initial condition”.

Let $P$, $Q$, and $\ds \frac{\partial Q}{\partial x}$ be continuous over some region of interest. Then \[ P\left( x, y\right)+ Q\left( x, y\right)\frac{dy}{dx}=0 \] is an exact ODE if and only if (iff) \[ \frac{\partial P}{\partial y} =\frac{\partial Q}{\partial x} \] everywhere in the region.

Proof: $\nec$ Recall that $\ds \frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}$ for a smooth function $\,f$. So if the ODE is exact, then there exists $\,f$ such that \[ f_x = \frac{\partial f}{\partial x} = P\left(x,y\right)\quad \text{and}\quad f_y = \frac{\partial f}{\partial y} = Q\left(x,y\right) \]

It follows that

Proof: $\suf\;$ If $\ds \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x},$ then we can always find $\,f$ such that $f_x=P\left(x,y\right)$ and $f_y=Q\left(x,y\right)$. Indeed, let \[ f\left(x,y\right)=\int_{x_0}^{x}P\left(s,y\right)ds + \int_{y_0}^{y}Q\left(x_0,t\right)dt \] for fixed $x_0,y_0$.

\[ \text{👉}\quad f\left(x,y\right)=\int_{x_0}^{x}P\left(s,y\right)ds + \int_{y_0}^{y}Q\left(x_0,t\right)dt \,\text{ for fixed }\, x_0,y_0. \]

For this proof we used the function \[ f\left(x,y\right)=\int_{x_0}^{x}P\left(s,y\right)ds + \int_{y_0}^{y}Q\left(x_0,t\right)dt. \]

Here we have that $P = 2x+ e^y$ and $Q = xe^y$.

So $\ds \frac{\partial P}{\partial y}$ $= e^y$ $ = \ds \frac{\partial Q}{\partial x}$. This means it is an exact ODE.

So there exists $\,f$ such that

Compare with $(2)$ above to obtain $\ds \frac{dg}{dy} = 0 $ $\;\Ra \;g(y) = C.$

Compare with $(2)$ above to obtain $\ds \frac{dg}{dy} = 0 $ $\;\Ra \;g(y) = C.$

And since $f = x^2 + xe^y + g(y),$ then \[ f\left(x, y\right) = x^2 + xe^y + C. \]

This is an implicit solution: $x^2 + xe^y = K.$

You should check that computing the derivative in both sides, implies that the ODE is satisfied. 📝

Let's say that we have an equation \[ P\left(x,y\right) + Q\left(x,y\right) \frac{dy}{dx} = 0 \] such that $\ds \frac{\partial P}{\partial y}\neq \frac{\partial Q}{\partial x}. $

The test we have just seen tells us that the ODE is not exact. Are we still able to do anything with it? Here we consider using an “integrating factor”, which is different to the one introduced to solve linear ODEs.

The idea is to multiply the ODE by a function $h(x, y)$ and then see if it is possible to choose $h(x, y)$ such that the resulting equation \[ h\left(x,y\right) P\left(x,y\right) + h\left(x,y\right) Q\left(x,y\right) \frac{dy}{dx} = 0 \] is exact. We know from the test that this new equation is exact if and only if \[ \frac{\partial }{\partial y}\big(h P \big) = \frac{\partial }{\partial x}\big(h Q \big). \]

Let's see if we can find such a function:

Let's see if we can find such a function:

From $\ds \frac{\partial }{\partial y}\big(h P \big) = \frac{\partial }{\partial x}\big(h Q \big)$, we have that

In general, the equation for $h(x, y)$ is usually just as difficult to solve as the original ODE. In some cases, however, we may be able to find an integrating factor which is a function of only one of the variables $x$ or $y$.

Let's try $h \equiv h(x)$:

Then we have that \[ \frac{dh}{dx} = h \left(\frac{P_y-Q_x}{Q}\right). \]

If more over $\ds \frac{P_y-Q_x}{Q}$ is a function of $x$ only, then this is a separable ODE and can be solved for $h(x)$.

Similarly for $h\equiv h(y)$.

Here we have that $\ds \frac{\partial P }{\partial y} = 3x+2y$ and $\ds \frac{\partial Q}{\partial x} =2x+y$.

Then $\ds \frac{\partial P }{\partial y} \neq \frac{\partial Q}{\partial x} $ Thus, it is not exact! 😮   It seems to be almost exact?

Try for example $h\equiv h(x)\neq 0$, then $$\ds h\left(3xy+y^2\right)+h\left(x^2+xy\right)\frac{dy}{dx}=0$$

Try for example $h\equiv h(x)\neq 0$, then $$\ds \overbrace{h\left(3xy+y^2\right)}^{{\large \hat P }}+\overbrace{h\left(x^2+xy\right)}^{{\large \hat Q}}\frac{dy}{dx}=0$$

is exact iff $\ds \frac{\partial \hat P }{\partial y} = \frac{\partial \hat Q}{\partial x}.$

Then $\ds \frac{\partial \hat P}{\partial y} = h\left(3x+2y\right)$ $\ds =\frac{\partial \hat Q}{\partial x}$ $=h\left(2x+y\right) +h'\left(x^2+xy\right)$

Thus $h\left(x+y\right)$ $=h'x \left(x+y\right)$   or   $h = h' x$   $\Ra \;h' = \ds \frac{h}{x}$   $\Ra \;h = K x$ for any constant $K$. Choose $K=1$ to obtain a new ODE: \[ 3x^2y+xy^2+\left(x^3+x^2y\right)\frac{dy}{dx}=0 \]

\[ \text{Exact ODE  👉   }\;3x^2y+xy^2+\left(x^3+x^2y\right)\frac{dy}{dx}=0 \] Note: its solution is given by $\ds x^3y+\frac{1}{2}x^2y^2=C.$ 📝