How to find the solution to an exact differential equation

Exact differential equations have a specific format, and are solved using a specific set of steps

In order for a differential equation to be called an exact differential equation, it must be given in the form

$M(x,y)+N(x,y)\frac{dy}{dx}=0$

and have a solution $\Psi(x,y)$ such that

$\frac{\partial\Psi}{\partial x}=\Psi_x=M(x,y)=M$

and

$\frac{\partial\Psi}{\partial y}=\Psi_y=N(x,y)=N$

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If we assume that $\Psi(x,y)$ is a continuous function, then we can also assume that its mixed second-order partial derivatives are equal.

$\frac{\partial^2\Psi}{\partial x\partial y}=\frac{\partial^2\Psi}{\partial y\partial x}$

We can rewrite this equation in a different form as

$\Psi_{xy}=\Psi_{yx}$

$\left(\Psi_x\right)_y=\left(\Psi_y\right)_x$

$(M)_y=(N)_x$

$M_y=N_x$

In other words, we just used the fact that $\Psi(x,y)$ has to be a solution to the exact differential equation to develop a test for exact differential equations, and the test tells us that a differential equation is exact if $M_y=N_x$. So, if the partial derivative of $M$ with respect to $y$ is equal to the partial derivative of $N$ with respect to $x$, then the differential equation is exact.

This is a test we can use to say whether or not the differential equation is exact, before we go about finding the solution $\Psi(x,y)$.

The general or implicit solution to an exact differential equation is given by

$\Psi(x,y)=c$

where $c$ is a constant. If we want to, we can prove that this is the solution by starting with the standard form of an exact differential equation

$M(x,y)+N(x,y)\frac{dy}{dx}=0$

We already said that

$\frac{\partial\Psi}{\partial x}=\Psi_x=M(x,y)=M$

and

$\frac{\partial\Psi}{\partial y}=\Psi_y=N(x,y)=N$

so we can substitute into the standard form and get

$\Psi_x+\Psi_y\frac{dy}{dx}=0$

Using chain rule for multivariable functions, we can change the left-hand side to

$\frac{d}{dx}\left[\Psi(x,y)\right]=0$

Integrating both sides with respect to $x$ to get $\Psi(x,y)$ by itself, we get

$\int\frac{d}{dx}\left[\Psi(x,y)\right]\ dx=\int 0\ dx$

$\Psi(x,y)+c_1=c_2$

$\Psi(x,y)=c_2-c_1$

$\Psi(x,y)=c$

Now that we’ve proven that $\Psi(x,y)=c$ is always the solution to an exact differential equation, we need to work on finding $\Psi(x,y)$.

We’ve already used the fact that

$\frac{\partial\Psi}{\partial x}=\Psi_x=M(x,y)=M$

and

$\frac{\partial\Psi}{\partial y}=\Psi_y=N(x,y)=N$

and we’ll use it again here to find the solution. We’ll start with

$\Psi_x=M(x,y)$

and

$\Psi_y=N(x,y)$

The left-hand sides of both of these are partial derivatives of $\Psi$, but we need to get back to $\Psi$ itself. To do that, we’d need to take the integrals

$\int\Psi_x\ dx=\int M(x,y)\ dx$

and

$\int\Psi_y\ dy=\int N(x,y)\ dy$

Taking the integral with respect to $x$ of the partial derivative with respect to $x$ cancels both operations and leaves us with just $\Psi$, in the same way that taking the integral with respect to $y$ of the partial derivative with respect to $y$ cancels those operations and leaves us with just $\Psi$.

$\Psi=\int M(x,y)\ dx$

and

$\Psi=\int N(x,y)\ dy$

In other words, if we want to find an equation for $\Psi$, we can either take the integral of $M$ with respect to $x$, or the integral of $N$ with respect to $y$. Both integrals will work, so we should look at $M$ and $N$ and then choose whichever function will be easier to integrate.

If we use the first integral, the one with $M$, we have to remember that we’re integrating a multivariable function in terms of $x$ and $y$ with respect to $x$ only. Which means that, instead of adding $+C$ to account for the constant of integration after we integrate, we’ll have to add $+h(y)$ to account for a function in terms of $y$.

Similarly, if we use the second integral, the one with $N$, we have to remember that we’re integrating a multivariable function in terms of $x$ and $y$ with respect to $y$ only. Which means that, instead of adding $+C$ to account for the constant of integration after we integrate, we’ll have to add $+h(x)$ to account for a function in terms of $x$.

Then it’s just a matter of solving for $h(y)$ or $h(x)$, which we’ll do by differentiating the equation for $\Psi$ with respect to $y$ if we’re trying to find $h(y)$, or with respect to $x$ if we’re trying to find $h(x)$.

That differentiation process will give us either $\Psi_x$ with $h\prime(x)$ or $\Psi_y$ with $h\prime(y)$. We’ll substitute $M(x,y)$ for $\Psi_x$ or $N(x,y)$ for $\Psi_y$, and then simplify the equation to solve for $h\prime(y)$ or $h\prime(x)$. Then we can integrate both sides of the remaining equation to solve for $h(y)$ or $h(x)$.

Finally, we’ll plug $h(y)$ or $h(x)$ back into the equation for $\Psi$, set the equation equal to $c$, and this will be the general or implicit solution to the exact differential equation.

In summary, to find the solution to an exact differential equation, we’ll

1. Verify that $M_y=N_x$ to confirm the differential equation is exact.

2. Use $\Psi=\int M(x,y)\ dx$ or $\Psi=\int N(x,y)\ dy$ to find $\Psi(x,y)$, including a value for $h(y)$ or $h(x)$.

3. Set $\Psi(x,y)=c$ to get the implicit solution.

How to solve an exact differential equation

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Step-by-step solution for finding the solution to the exact differential equation

Example

If the differential equation is exact, find its solution.

$x^3y^4+(x^4y^3+2y)\frac{dy}{dx}=0$

First, we’ll test to see whether or not the differential equation is exact. Matching the given equation to the standard form of an exact differential equation, we can say that

$M(x,y)=x^3y^4$

and

$N(x,y)=x^4y^3+2y$

We’ll test to see whether $M_y=N_x$.

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

$4x^3y^3=4x^3y^3$

Both integrals will work, so we should look at M and N and then choose whichever function will be easier to integrate.

Since $M_y=N_x$, we know that the given differential equation is an exact differential equation. Now we just need to find the solution $\Psi(x,y)=c$.

$M(x,y)$ and $N(x,y)$ are equally easy to integrate, we’ll just use $M(x,y)$ and the integral

$\Psi=\int M(x,y)\ dx$

$\Psi=\int x^3y^4\ dx$

$\Psi=\frac14x^4y^4+h(y)$

We had to add $h(y)$ instead of just $C$ because we integrated a multivariable function with respect to $x$ only, which doesn’t account for the integration of $y$. Now we need to find $h(y)$, which we’ll do by taking the partial derivative of both sides of the equation with respect to $y$.

$\frac{d}{dy}\Psi=\frac{d}{dy}\left[\frac14x^4y^4+h(y)\right]$

$\Psi_y=x^4y^3+h\prime(y)$

Because we know that $\Psi_y=N(x,y)$, we’ll make that substitution and then solve for $h\prime(y)$.

$x^4y^3+2y=x^4y^3+h\prime(y)$

$2y=h\prime(y)$

To find $h(y)$, we’ll integrate both sides with respect to $y$.

$\int2y\ dy=\int h\prime(y)\ dy$

$y^2+k_1=h(y)+k_2$

$h(y)=y^2+k_1-k_2$

$h(y)=y^2+k$

Plugging this into the equation for $\Psi$ gives

$\Psi=\frac14x^4y^4+h(y)$

$\Psi=\frac14x^4y^4+y^2+k$

Finally, setting $\Psi=c$ to find the solution to the exact differential equation, we get

$c=\frac14x^4y^4+y^2+k$

$c-k=\frac14x^4y^4+y^2$

$c=\frac14x^4y^4+y^2$

We can say that the equation is exact and that its general (implicit) solution is

$c=\frac14x^4y^4+y^2$

Sometimes we’ll be given an initial condition and asked to find an explicit solution, instead of a general (implicit) solution. If that’s the case, we just plug the initial condition into the solution to find a value for $c$. For example, if the previous example had given the initial condition $y(0)=2$, our solution would have been

$c=\frac14x^4y^4+y^2$

$c=\frac14(0)^4(2)^4+(2)^2$

$c=4$

$4=\frac14x^4y^4+y^2$

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