# Solving Exact Differential Equations Examples 1

Recall from the Exact Differential Equations page that a differential equation of the form $M(x, y) + N(x, y) \frac{dy}{dx} = 0$ is said to be exact if there exists a function $\psi (x, y)$ such that $\frac{\partial}{\partial x} \psi (x, y) = M(x, y)$ and $\frac{\partial}{\partial y} \psi (x, y) = N(x, y)$.

Furthermore, we noted that a differential is exact if and only if $\frac{\partial}{\partial y} M(x, y) = \frac{\partial}{\partial x} N(x, y)$ (and under some conditions on the continuity of $M$, $N$, $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$.

We will now look at some more examples of solving exact differential equations.

## Example 1

**Show that the differential equation $\frac{dy}{dx} = – \frac{Ax + By}{Bx + Cy}$ for $A, B, C \in \mathbb{R}$ is exact and solve this differential equation.**

We first rewrite this differential equation in the form $M(x, y) + N(x, y) \frac{dy}{dx} = 0$:

(1)

We have that $M(x, y) = Ax + By$ and $N(x, y) = Bx + Cy$. The partial derivative of $M$ with respect to $y$ and the partial derivative of $N$ with respect to $x$ are:

(2)

Indeed this differential equation is exact, and so there exists a function $\psi (x, y)$ such that:

(3)

From the first equation, $\frac{\partial \psi}{\partial x} = Ax + By$, we integrate with respect to $x$ and get:

(4)

We now partial differentiate with respect to $y$ and get:

(5)

We are already given that $\frac{\partial \psi}{\partial y} = N(x, y) = Bx + Cy$. Therefore $h'(y) = Cy$ and so $h(y) = \int Cy \: dy = \frac{Cy^2}{2}$. Therefore the solution to our differential equation is:

(6)

## Example 2

**Show that the differential equation $\frac{dy}{dt} = \frac{- \left ( ye^{xy} \cos 2x – 2e^{xy} \sin 2x + 2x \right )}{(xe^{xy} \cos 2x – 3)}$ is an exact and solve this differential equation.**

We first rewrite this differential equation in the form $M(x, y) + N(x, y) \frac{dy}{dx} = 0$:

(7)

Let $M(x, y) = ye^{xy} \cos 2x – 2e^{xy} \sin 2x + 2x$ and let $N(x, y) = xe^{xy} \cos 2x – 3$. The partial derivative of $M$ with respect to $y$ and the partial derivative of $N$ with respect to $x$ are:

(8)

Therefore the given differential equation is indeed exact, and so there exists a function $\psi (x, y)$ such that:

(9)

We now take the second equation and integrate with respect to $y$ (this is much simpler than integrating the first equation with respect to $x$). We have that:

(10)

We now partial differentiate both sides with respect to $x$ to get that:

(11)

We already note that $\frac{\partial \psi}{\partial x} = M(x, y) = ye^{xy} \cos 2x – 2e^{xy} \sin 2x + 2x$. By comparing this with above, we see that $h'(x) = 2x$, so $h(x) = x^2$, and so the solution to this differential equation is:

(12)