# Exact Equations

The differential equation ${M}{\left({x},{y}\right)}{d}{x}+{N}{\left({x},{y}\right)}{d}{y}={0}$ is exact, if there exists a function ${f{}}$ such that ${d}{f{=}}{M}{\left({x},{y}\right)}{d}{x}+{N}{\left({x},{y}\right)}{d}{y}$.

In this case, the equation can be rewritten as ${d}{f{=}}{0}$, which gives the solution ${f{=}}{C}$.

Test for exactness: if ${M}{\left({x},{y}\right)}$ and ${N}{\left({x},{y}\right)}$ are continuous functions and have continuous first partial derivatives on some rectangle of the xy-plane, the differential equation is exact, if and only if $\frac{{\partial{M}{\left({x},{y}\right)}}}{{\partial{y}}}=\frac{{\partial{N}{\left({x},{y}\right)}}}{{\partial{x}}}$.

To solve this equation, we use the facts that $\frac{{\partial{f{{\left({x},{y}\right)}}}}}{{\partial{x}}}={M}{\left({x},{y}\right)}$ and $\frac{{\partial{f{{\left({x},{y}\right)}}}}}{{\partial{y}}}={N}{\left({x},{y}\right)}$. We integrate the first equation with respect to ${x}$ to obtain ${f{{\left({x},{y}\right)}}}$ through ${x}$ and an unknown function ${g{{\left({y}\right)}}}$. We then differentiate the result with respect to ${y}$ and use the second equation. After this, we find ${g{{\left({y}\right)}}}$ and thus ${f{{\left({x},{y}\right)}}}$. The solution, as already stated above, is given as ${f{{\left({x},{y}\right)}}}={C}$.

Example 1. Solve ${\left({x}+{\sin{{\left({y}\right)}}}\right)}{d}{x}+{\left({x}{\cos{{\left({y}\right)}}}-{2}{y}\right)}{d}{y}={0}$.

Here, ${M}{\left({x},{y}\right)}={x}+{\sin{{\left({y}\right)}}}$, ${N}{\left({x},{y}\right)}={x}{\cos{{\left({y}\right)}}}-{2}{y}$. Since $\frac{{\partial{M}}}{{\partial{y}}}={\cos{{\left({y}\right)}}}$ and $\frac{{\partial{N}}}{{\partial{x}}}={\cos{{\left({y}\right)}}}$, we have that $\frac{{\partial{M}}}{{\partial{y}}}=\frac{{\partial{N}}}{{\partial{x}}}$ and the differential equation is exact.

Thus, there exists a function ${f{}}$ such that $\frac{{\partial{f}}}{{\partial{x}}}={M}{\left({x},{y}\right)}={x}+{\sin{{\left({y}\right)}}}$ and $\frac{{\partial{f}}}{{\partial{y}}}={N}{\left({x},{y}\right)}={x}{\cos{{\left({y}\right)}}}-{2}{y}$.

Integrate the first equation with respect to ${x}$ to obtain that ${f{=}}\int{\left({x}+{\sin{{\left({y}\right)}}}\right)}{d}{x}=\frac{{1}}{{2}}{{x}}^{{2}}+{x}{\sin{{\left({y}\right)}}}+{g{{\left({y}\right)}}}$.

Note that here, the constant of integration can depend on ${y}$ (because we integrate with respect to ${x}$).

Now, differentiate the resulting equation with respect to ${y}$:

$\frac{{\partial{f}}}{{\partial{y}}}={0}+{x}{\cos{{\left({y}\right)}}}+{g{'}}{\left({y}\right)}$.

On the other hand, $\frac{{\partial{f}}}{{\partial{y}}}={N}{\left({x},{y}\right)}={x}{\cos{{\left({y}\right)}}}-{2}{y}$.

So, ${x}{\cos{{\left({y}\right)}}}+{g{'}}{\left({y}\right)}={x}{\cos{{\left({y}\right)}}}-{2}{y}$, or ${g{'}}{\left({y}\right)}=-{2}{y}$. Integrating it, we obtain that ${g{{\left({y}\right)}}}=-{{y}}^{{2}}+{c}_{{1}}$.

Thus,

${f{=}}\frac{{1}}{{2}}{{x}}^{{2}}+{x}{\sin{{\left({y}\right)}}}-{{y}}^{{2}}+{c}_{{1}}$. The solution is given as $\frac{{1}}{{2}}{{x}}^{{2}}+{x}{\sin{{\left({y}\right)}}}-{{y}}^{{2}}+{c}_{{1}}={C}$,

Or $\frac{{1}}{{2}}{{x}}^{{2}}+{x}{\sin{{\left({y}\right)}}}-{{y}}^{{2}}={c}_{{2}}$, where ${c}_{{2}}={C}-{c}_{{1}}$.

We obtain an implicit solution, and there is no way to find an explicit solution.

Let's work another example.

Example 2. Solve ${y}'=\frac{{{2}+{y}{{e}}^{{{x}{y}}}}}{{{2}{y}-{x}{{e}}^{{{x}{y}}}}}$.

First, rewrite it into the differential form: $\frac{{{d}{y}}}{{{d}{x}}}=\frac{{{2}+{y}{{e}}^{{{x}{y}}}}}{{{2}{y}-{x}{{e}}^{{{x}{y}}}}}$, or ${\left({2}+{y}{{e}}^{{{x}{y}}}\right)}{d}{x}+{\left({x}{{e}}^{{{x}{y}}}-{2}{y}\right)}{d}{y}={0}$.

Here, ${M}{\left({x},{y}\right)}={2}+{y}{{e}}^{{{x}{y}}}$, and ${N}{\left({x},{y}\right)}={x}{{e}}^{{{x}{y}}}-{2}{y}$.

Since $\frac{{\partial{M}}}{{\partial{y}}}={{e}}^{{{x}{y}}}+{x}{y}{{e}}^{{{x}{y}}}$ and $\frac{{\partial{N}}}{{\partial{x}}}={{e}}^{{{x}{y}}}+{x}{y}{{e}}^{{{x}{y}}}$, we have that$\frac{{\partial{M}}}{{\partial{y}}}=\frac{{\partial{N}}}{{\partial{x}}}$ and the differential equation is exact.

So, there exists a function ${f{}}$ such that $\frac{{\partial{f}}}{{\partial{x}}}={M}{\left({x},{y}\right)}={2}+{y}{{e}}^{{{x}{y}}}$ and $\frac{{\partial{f}}}{{\partial{y}}}={N}{\left({x},{y}\right)}={x}{{e}}^{{{x}{y}}}-{2}{y}$. In the previous example, we took the first equation and integrated it with respect to ${x}$. Actually, we could take the second equation and integrate it with respect to ${y}$. The final answer is the same. In this example, we will take the second equation and integrate it with respect to ${y}$:

${f{=}}\int{\left({x}{{e}}^{{{x}{y}}}-{2}{y}\right)}{d}{y}={{e}}^{{{x}{y}}}-{{y}}^{{2}}+{g{{\left({x}\right)}}}$. Note that the constant of integration depends on ${x}$, since we integrate with respect to ${y}$.

Now, differentiate the resulting equation with respect to ${x}$:

$\frac{{\partial{f}}}{{\partial{x}}}={y}{{e}}^{{{x}{y}}}+{g{'}}{\left({x}\right)}$.

On the other hand, $\frac{{\partial{f}}}{{\partial{x}}}={2}+{y}{{e}}^{{{x}{y}}}$; so, ${y}{{e}}^{{{x}{y}}}+{g{'}}{\left({x}\right)}={2}+{y}{{e}}^{{{x}{y}}}$, or ${g{'}}{\left({x}\right)}={2}$.

Integrating with respect to ${x}$ gives ${g{{\left({x}\right)}}}={2}{x}+{c}_{{1}}$.

So, ${f{=}}{{e}}^{{{x}{y}}}-{{y}}^{{2}}+{g{{\left({x}\right)}}}={{e}}^{{{x}{y}}}-{{y}}^{{2}}+{2}{x}+{c}_{{1}}$.

And the solution is ${{e}}^{{{x}{y}}}-{{y}}^{{2}}+{2}{x}+{c}_{{1}}={C}$ or ${{e}}^{{{x}{y}}}-{{y}}^{{2}}+{2}{x}={c}_{{2}}$, where ${c}_{{2}}={C}-{c}_{{1}}$.

Sometimes, an equation can be not exact, but it can be transformed into an exact one by multiplying the equation by the integrating factor.

So, ${I}{\left({x},{y}\right)}$ is the integrating factor for the differential equation, if ${I}{\left({x},{y}\right)}{\left({M}{\left({x},{y}\right)}{d}{x}+{N}{\left({x},{y}\right)}{d}{y}\right)}={0}$ is exact.

There are 2 conditions that allow to find the integrating factor easily:

If $\frac{{1}}{{N}}{\left(\frac{{\partial{M}}}{{\partial{y}}}-\frac{{\partial{N}}}{{\partial{x}}}\right)}={g{{\left({x}\right)}}}$, a function of ${x}$ alone, then ${I}{\left({x},{y}\right)}={{e}}^{{\int{g{{\left({x}\right)}}}{d}{x}}}$.

If $\frac{{1}}{{M}}{\left(\frac{{\partial{M}}}{{\partial{y}}}-\frac{{\partial{N}}}{{\partial{x}}}\right)}={h}{\left({y}\right)}$, a function of ${y}$ alone, then ${I}{\left({x},{y}\right)}={{e}}^{{-\int{h}{\left({y}\right)}{d}{y}}}$.

Now, let's do some more work with another example

Example 3. Solve ${\left({y}+{1}\right)}{d}{x}-{x}{d}{y}={0}$.

Here, ${M}{\left({x},{y}\right)}={y}+{1}$, and ${N}{\left({x},{y}\right)}=-{x}$. Since $\frac{{\partial{M}}}{{\partial{y}}}={1}$ and $\frac{{\partial{N}}}{{\partial{x}}}=-{1}$, we have that $\frac{{\partial{M}}}{{\partial{y}}}\ne\frac{{\partial{N}}}{{\partial{x}}}$ and the equation is not exact. However, note that $\frac{{1}}{{N}}{\left(\frac{{\partial{M}}}{{\partial{y}}}-\frac{{\partial{N}}}{{\partial{x}}}\right)}=\frac{{1}}{{-{x}}}{\left({1}-{\left(-{1}\right)}\right)}=-\frac{{2}}{{x}}={g{{\left({x}\right)}}}$.

So, ${I}{\left({x},{y}\right)}={{e}}^{{\int-\frac{{2}}{{x}}{d}{x}}}={{e}}^{{-{2}{\ln{{\left({x}\right)}}}}}=\frac{{1}}{{{x}}^{{2}}}$.

Multiplying the differential equation by the integrating factor gives: $\frac{{{y}+{1}}}{{{x}}^{{2}}}{d}{x}-\frac{{1}}{{x}}{d}{y}={0}$.

This equation is exact; so, there exists a function ${f{}}$ such that ${d}{f{=}}\frac{{{y}+{1}}}{{{x}}^{{2}}}{d}{x}-\frac{{1}}{{x}}{d}{y}$. Using the fact that $\frac{{\partial{f}}}{{\partial{x}}}=\frac{{{y}+{1}}}{{{x}}^{{2}}}$, we find that ${f{=}}\int\frac{{{y}+{1}}}{{{x}}^{{2}}}{d}{x}=-\frac{{{y}+{1}}}{{x}}+{g{{\left({y}\right)}}}$. Now, differentiating with respect to ${y}$ gives: $\frac{{\partial{f}}}{{\partial{y}}}=-\frac{{1}}{{x}}+{g{'}}{\left({y}\right)}$. On the other hand, $\frac{{\partial{f}}}{{\partial{y}}}=-\frac{{1}}{{x}}$; so, $-\frac{{1}}{{x}}+{g{'}}{\left({y}\right)}=-\frac{{1}}{{x}}$, or ${g{'}}{\left({y}\right)}={0}$.

Solving it, we obtain that ${g{{\left({y}\right)}}}={C}_{{1}}$.

So, ${f{{\left({x},{y}\right)}}}=-\frac{{{y}+{1}}}{{x}}+{g{{\left({y}\right)}}}=-\frac{{{y}+{1}}}{{x}}+{C}_{{1}}$.

Therefore, the solution is $-\frac{{{y}+{1}}}{{x}}+{C}_{{1}}={C}$, or ${y}={c}_{{2}}{x}-{1}$, where ${c}_{{2}}={C}_{{1}}-{C}$.

Let's solve another equation.

Example 4. Solve ${2}{x}{y}{d}{x}+{{y}}^{{2}}{d}{y}={0}$.

Here, ${M}{\left({x},{y}\right)}={2}{x}{y}$, and ${N}{\left({x},{y}\right)}={{y}}^{{2}}$. Since $\frac{{\partial{M}}}{{\partial{y}}}={2}{x}$ and $\frac{{\partial{N}}}{{\partial{x}}}={0}$, we have that $\frac{{\partial{M}}}{{\partial{y}}}\ne\frac{{\partial{N}}}{{\partial{x}}}$ and the differential equation is not exact.

However, note that $\frac{{1}}{{M}}{\left(\frac{{\partial{M}}}{{\partial{y}}}=\frac{{\partial{N}}}{{\partial{x}}}\right)}=\frac{{1}}{{{2}{x}{y}}}{\left({2}{x}-{0}\right)}=\frac{{1}}{{y}}={h}{\left({y}\right)}$.

So, the integrating factor is ${I}{\left({x},{y}\right)}={{e}}^{{-\int\frac{{1}}{{y}}{d}{y}}}={{e}}^{{-{\ln{{\left({y}\right)}}}}}=\frac{{1}}{{y}}$.

Multiplying the differential equation by the integrating factor yields $\frac{{1}}{{y}}{\left({2}{x}{y}{d}{x}+{{y}}^{{2}}{d}{y}\right)}={0}$, or ${2}{x}{d}{x}+{y}{d}{y}={0}$.

This equation is exact. Using the fact that $\frac{{\partial{f}}}{{\partial{x}}}={2}{x}$, we have that ${f{=}}\int{\left({2}{x}\right)}{d}{x}={{x}}^{{2}}+{h}{\left({y}\right)}$. Differentiating the last eqaution with respect to ${y}$ gives: $\frac{{\partial{f}}}{{\partial{y}}}={h}'{\left({y}\right)}$.

On the other hand, $\frac{{\partial{f}}}{{\partial{y}}}={y}$; so, ${h}'{\left({y}\right)}={y}$, or ${h}{\left({y}\right)}=\int{y}{d}{y}=\frac{{1}}{{2}}{{y}}^{{2}}+{c}_{{1}}$.

Hence, ${f{=}}{{x}}^{{2}}+{h}{\left({y}\right)}={{x}}^{{2}}+\frac{{1}}{{2}}{{y}}^{{2}}+{c}_{{1}}$, and the solution is ${{x}}^{{2}}+\frac{{1}}{{2}}{{y}}^{{2}}+{c}_{{1}}={C}$, or ${{x}}^{{2}}+\frac{{1}}{{2}}{{y}}^{{2}}={c}_{{2}}$, where ${c}_{{2}}={C}-{c}_{{1}}$.

Another case when the integrating factor can be found easily is when ${M}={y}{f{{\left({x}{y}\right)}}}$ and ${N}={x}{g{{\left({x}{y}\right)}}}$. In this case, ${I}{\left({x},{y}\right)}=\frac{{1}}{{{x}{M}-{y}{N}}}$.

Example 5. Solve ${y}{\left({1}-{x}{y}\right)}{d}{x}+{x}{d}{y}={0}$.

The equation is not exact; however, note that ${M}{\left({x},{y}\right)}$ is in the form ${y}{\left({1}-{x}{y}\right)}$ and ${N}{\left({x},{y}\right)}={x}\cdot{1}$; so, the integrating factor is ${I}{\left({x},{y}\right)}=\frac{{1}}{{{x}{y}{\left({1}-{x}{y}\right)}-{x}{y}}}=-\frac{{1}}{{{\left({x}{y}\right)}}^{{2}}}$. Multiplying by ${I}{\left({x},{y}\right)}$ yields:

$\frac{{{x}{y}-{1}}}{{{{x}}^{{2}}{y}}}{d}{x}-\frac{{1}}{{{x}{{y}}^{{2}}}}{d}{y}={0}$. This equation is exact.

Using the fact that $\frac{{\partial{f}}}{{\partial{x}}}=\frac{{{x}{y}-{1}}}{{{{x}}^{{2}}{y}}}$, we have that ${f{=}}\int{\left(\frac{{{x}{y}-{1}}}{{{{x}}^{{2}}{y}}}\right)}{d}{x}={\ln{{\left({\left|{x}\right|}\right)}}}+\frac{{1}}{{{x}{y}}}+{g{{\left({y}\right)}}}$.

Differentiating with respect to ${y}$ gives: $\frac{{\partial{f}}}{{\partial{y}}}=-\frac{{1}}{{{x}{{y}}^{{2}}}}+{g{'}}{\left({y}\right)}$. On the other hand, $\frac{{\partial{f}}}{{\partial{y}}}=-\frac{{1}}{{{x}{{y}}^{{2}}}}$.

So, ${g{'}}{\left({y}\right)}={0}$, or ${g{{\left({y}\right)}}}={c}_{{1}}$.

Thus, ${f{{\left({x},{y}\right)}}}={\ln{{\left({\left|{x}\right|}\right)}}}+\frac{{1}}{{{x}{y}}}+{c}_{{1}}$.

Finally, the solution is ${\ln{{\left({\left|{x}\right|}\right)}}}+\frac{{1}}{{{x}{y}}}+{c}_{{1}}={C}$, or ${\ln{{\left({\left|{x}\right|}\right)}}}+\frac{{1}}{{{x}{y}}}={c}_{{2}}$, where ${c}_{{2}}={C}-{c}_{{1}}$.

Below is the list of common integrating factors:

 Group of terms ${I}{\left({x},{y}\right)}$ Exact differential ${d}{g{{\left({x},{y}\right)}}}$ ${y}{d}{x}-{x}{d}{y}$ $-\frac{{1}}{{{x}}^{{2}}}$ $-\frac{{{y}{d}{x}-{x}{d}{y}}}{{{x}}^{{2}}}={d}{\left(\frac{{y}}{{x}}\right)}$ ${y}{d}{x}-{x}{d}{y}$ $\frac{{1}}{{{y}}^{{2}}}$ $\frac{{{y}{d}{x}-{x}{d}{y}}}{{{y}}^{{2}}}={d}{\left(\frac{{x}}{{y}}\right)}$ ${y}{d}{x}-{x}{d}{y}$ $-\frac{{1}}{{{x}{y}}}$ $-\frac{{{y}{d}{x}-{x}{d}{y}}}{{{x}{y}}}={d}{\left({\ln{{\left(\frac{{y}}{{x}}\right)}}}\right)}$ ${y}{d}{x}-{x}{d}{y}$ $-\frac{{1}}{{{{x}}^{{2}}+{{y}}^{{2}}}}$ $-\frac{{{y}{d}{x}-{x}{d}{y}}}{{{{x}}^{{2}}+{{y}}^{{2}}}}={d}{\left({\operatorname{arctan}{{\left(\frac{{y}}{{x}}\right)}}}\right)}$ ${y}{d}{y}+{x}{d}{x}$ $\frac{{1}}{{{x}{y}}}$ $\frac{{{y}{d}{y}+{x}{d}{x}}}{{{x}{y}}}={d}{\left({\ln{{\left({x}{y}\right)}}}\right)}$ ${y}{d}{y}+{x}{d}{x}$ $\frac{{1}}{{{\left({x}{y}\right)}}^{{n}}},{n}>{1}$ $\frac{{{y}{d}{y}+{x}{d}{x}}}{{{\left({x}{y}\right)}}^{{n}}}=-{d}{\left(\frac{{1}}{{{\left({n}-{1}\right)}{{\left({x}{y}\right)}}^{{{n}-{1}}}}}\right)}$ ${y}{d}{y}+{x}{d}{x}$ $\frac{{1}}{{{{x}}^{{2}}+{{y}}^{{2}}}}$ $\frac{{{y}{d}{y}+{x}{d}{x}}}{{{{x}}^{{2}}+{{y}}^{{2}}}}={d}{\left(\frac{{1}}{{2}}{\ln{{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}}}\right)}$ ${y}{d}{y}+{x}{d}{x}$ $\frac{{1}}{{{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}}^{{n}}},{n}>{1}$ $\frac{{{y}{d}{y}+{x}{d}{x}}}{{{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}}^{{n}}}=-{d}{\left(\frac{{1}}{{{2}{\left({n}-{1}\right)}{{\left({{x}}^{{2}}+{{y}}^{{2}}\right)}}^{{{n}-{1}}}}}\right)}$ ${a}{y}{d}{x}+{b}{x}{d}{y}$ (${a}$ and ${b}$ are constants) ${{x}}^{{{a}-{1}}}{{y}}^{{{b}-{1}}}$ ${{x}}^{{{a}-{1}}}{{y}}^{{{b}-{1}}}{\left({a}{y}{d}{x}+{b}{x}{d}{y}\right)}={d}{\left({{x}}^{{a}}{{y}}^{{b}}\right)}$

In general, integrating factors are difficult to uncover. If a differential equation doesn't have one of the forms given above, searching for the integrating factor will likely be unsuccessful, and one should resort to other solution methods.