Solving Exact Differential Equations Examples 2

# Solving Exact Differential Equations Examples 2

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 $\frac{x}{(x^2 + y^2)^{3/2}} + \frac{y}{(x^2 + y^2)^{3/2}} \frac{dy}{dx} = 0$ is exact and solve this differential equation.

Let $M(x, y) = \frac{x}{(x^2 + y^2)^{3/2}}$ and let $N(x, y) = \frac{y}{(x^2 + y^2)^{3/2}}$. The partial derivative of $M$ with respect to $y$ and the partial derivative of $N$ with respect to $x$ are:

(1)
\begin{align} \quad \frac{\partial M}{\partial y} = \frac{-3xy (x^2 + y^2)^{1/2}}{(x^2 + y^2)^3} = \frac{-3xy}{(x^2 + y^2)^{5/2}} \\ \quad \frac{\partial N}{\partial x} = \frac{-3xy (x^2 + y^2)^{1/2}}{(x^2 + y^2)^3} = \frac{-3xy}{(x^2 + y^2)^{5/2}} \end{align}

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

(2)
\begin{align} \quad \frac{\partial \psi}{\partial x} = \frac{x}{(x^2 + y^2)^{3/2}} \quad , \quad \frac{\partial \psi}{\partial y} = \frac{y}{(x^2 + y^2)^{3/2}} \end{align}

We integrate the first equation with respect to $x$ to get that:

(3)
\begin{align} \quad \psi (x, y) = \int \frac{x}{(x^2 + y^2)^{3/2}} \: dx \end{align}

We can evaluate this integral with substitution. Let $u = x^2 + y^2$. Then $du = 2x \: dx$ and so $\frac{1}{2} du = x \: dx$. Thus:

(4)
\begin{align} \quad \psi (x, y) = \frac{1}{2} \int \frac{1}{u^{3/2}} \: du \\ \quad \psi (x, y) = \frac{1}{2} \int u^{-3/2} \: du \\ \quad \psi (x, y) = -u^{-1/2} \\ \quad \psi (x, y) = - \frac{1}{(x^2 + y^2)^{1/2}} + h(y) \end{align}

We now partial differentiate this function with respect to $y$ to get that:

(5)
\begin{align} \quad \frac{\partial \psi}{\partial y} = - \frac{-y(x^2 + y^2)^{-1/2}}{x^2 + y^2} + h'(y) \\ \quad \frac{\partial \psi}{\partial y} = \frac{y}{(x^2 + y^2)^{3/2}} + h'(y) \end{align}

We already have that $\frac{\partial \psi}{\partial y} = \frac{y}{(x^2 + y^2)^{3/2}}$ and so $h'(y) = 0$ which implies that $h(y) = C$. Therefore the solution to this differential equation is:

(6)
\begin{align} \quad - \frac{1}{(x^2 + y^2)^{1/2}} = D \end{align}