Exact Equations

As remarked, there are no general methods to find a solution of (7.1.2). The EXACT EQUATIONS is yet another class of equations that can be easily solved. In this section, we introduce this concept.

Let $D$ be a region in $xy$ -plane and let $M$ and $N$ be real valued functions defined on $D.$ Consider an equation

$$M(x, y) + N(x, y) \frac{dy}{dx} = 0, \; (x, y) \in D.$$ (7.3.1)

In most of the books on Differential Equations, this equation is also written as

$$M(x, y) dx + N(x, y) dy = 0, \; (x, y) \in D.$$ (7.3.2)

DEFINITION 7.3.1 (Exact Equation)   The Equation (7.3.1) is called Exact if there exists a real valued twice continuously differentiable function $f: {\mathbb{R}}^2 {\longrightarrow}{\mathbb{R}}$ (or the domain is an open subset of ${\mathbb{R}}^2$ ) such that

$$\frac{\partial f}{\partial x} = M \; {\mbox{ and }} \; \frac{\partial f}{\partial y} = N.$$ (7.3.3)

Remark 7.3.2   If (7.3.1) is exact, then

$$\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dx} = \frac{df(x, y)}{dx} = 0.$$

This implies that $f(x,y) = c$ (where $c$ is a constant) is an implicit solution of (7.3.1). In other words, the left side of (7.3.1) is an exact differential.

EXAMPLE 7.3.3   The equation $y + x \frac{dy}{dx} = 0$ is an exact equation. Observe that in this example, $f(x,y) = x y.$

The proof of the next theorem is given in Appendix 14.6.2.

THEOREM 7.3.4   Let $M$ and $N$ be twice continuously differentiable function in a region $D.$ The Equation (7.3.1) is exact if and only if

$$\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}.$$ (7.3.4)

Note: If (7.3.1) or (7.3.2) is exact, then there is a function $f(x,y)$ satisfying $f(x,y) = c$ for some constant $c,$ such that

$$d( f(x, y) ) = M(x, y) dx + N(x,y) dy = 0.$$

Let us consider some examples, where Theorem 7.3.4 can be used to easily find the general solution.

EXAMPLE 7.3.5  

1. Solve
   $$2 x e^{y} + (x^2 e^{y} + \cos y \; ) \frac{dy}{dx} = 0.\;\;$$
   
   Solution: With the above notations, we have
   $$M = 2 x e^{y}, \; N = x^2 e^{y} + \cos y, \; \frac{\partial M}{\... ...y} = 2 x e^{y} \; {\mbox{ and }} \; \frac{\partial N}{\partial x} = 2 x e^{y}.$$
   Therefore, the given equation is exact. Hence, there exists a function $G(x,y)$ such that
   $$\frac{\partial G}{\partial x} = 2 x e^{y} \;\;{\mbox{ and }}\;\; \frac{\partial G}{\partial y} = x^2 e^{y} + \cos y.$$
   The first partial differentiation when integrated with respect to $x$ (assuming $y$ to be a constant) gives,
   $$G(x,y) = x^2 e^{y} + h(y).$$
   But then
   $$\frac{\partial G}{\partial y} = \frac{\partial (x^2 e^{y} + h(y))}{\partial y} = N$$
   implies $\frac{dh}{dy} = \cos y$ or $h(y) = \sin y + c$ where $c$ is an arbitrary constant. Thus, the general solution of the given equation is
   $$x^2 e^{y} + \sin y = c.$$
   The solution in this case is in implicit form.
2. Find values of $\ell$ and $m$ such that the equation
   $$\ell y^2 + m x y \frac{dy}{dx} = 0$$
   is exact. Also, find its general solution.
   Solution: In this example, we have
   $$M = \ell y^2, \; N = m x y, \; \; \frac{\partial M}{\partial y} = 2 \ell y\; {\mbox{ and }} \; \frac{\partial N}{\partial x} = m y.$$
   Hence for the given equation to be exact, $m = 2 \ell.$ With this condition on $\ell$ and $m,$ the equation reduces to
   $$\ell y^2 + 2 \ell x y \frac{dy}{dx} = 0.$$
   This equation is not meaningful if $\ell = 0.$ Thus, the above equation reduces to
   $$\frac{d}{dx}(x y^2) = 0$$
   whose solution is
   $$x y^2 = c$$
   for some arbitrary constant $c.$
3. Solve the equation
   $$(3 x^2 e^{y} - x^2 ) dx + (x^3 e^{y} + y^2) dy = 0.$$
   
   Solution: Here
   $$M = 3 x^2 e^{y} - x^2 \; {\mbox{ and }}\; N = x^3 e^{y} + y^2.$$
   Hence, $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 3 x^2 e^{y}.$ Thus the given equation is exact. Therefore,
   $$G(x, y) = \int (3 x^2 e^{y} - x^2) d x = x^3 e^{y} - \frac{x^3}{3} + h(y)$$
   (keeping $y$ as constant). To determine $h(y),$ we partially differentiate $G(x,y)$ with respect to $y$ and compare with $N$ to get $h(y) = \frac{y^3}{3}.$ Hence
   $$G(x,y) = x^3 e^{y} - \frac{x^3}{3} + \frac{y^3}{3} = c$$
   is the required implicit solution.