Exact differential equations

The total differential of $u(x,y)$ is

$$du = u_{x}dx + u_{y}dy$$

Thus if we know the values of $u_{x},u_{y}$, then we can find $du$. On the other hand, if we know the total differential $du$ of a function $u$, then we can determine the function $u$.

Example 1..9   The total differential of $u(x,y)$ is given by

$$du = (2xy - \cos{x}dx + (x^{2} - 1)dy$$

Find $u(x,y)$.

SOLUTION Since $du = u_x dx + u_y dy$, we have

$$u_{x} = 2xy - \cos{x}, \ u_{y} = x^{2} - 1.$$

Integrate the first equation by $x$ to obtain

$$u(x,y) = \int( 2xy - \cos{x})dx = x^{2}y - \sin{x} + c(y)$$

Note that $c(y)$ is an arbitrary function of $y$. Now differentiate with respect ot $y$,

$$u_{y} = x^2 + c^{\prime}(y)$$

Now $u_{y}$ obtained and the $u_{y}$ given above must be the same. Thus

$$u_{y} = x^{2} + c^{\prime}(y) = x^{2} - 1 .$$

From this we have $c^{\prime}(y) = -1$ and $c(y) = -y + C$. Therefore,

$$u(x,y) = x^{2}y - \sin{x} - y + C\ensuremath{\ \blacksquare}$$

Definition 1..1   The left-hand side of the 1st order differential equation

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

is the same as the differential $du$ of some function $u$. Then the general solution is given by

$$u(x,y) = c$$

Example 1..10   Find the general solution of $(2xy - \cos{x}dx + (x^{2} - 1)dy = 0$.

SOLUTION In the example above, we found the function $u$ so that $du$ is equal to the left-hand side of equation. Thus the general solution is

$$x^{2}y - \sin{x} - y = c \ensuremath{\ \blacksquare}$$

Theorem 1..1   Let $M(x,y)$ and $N(x,y)$ be the class $C^{1}$ on $\{(x,y) : a < x < b, c < y < d \}$. Then the followings are equivalent
$(1) \ M(x,y)dx + N(x,y)dy = 0\$   is exact
$(2) \ M_{y} = N_{x}$

Proof $(1) \Rightarrow (2)$
If the differential equation is exact, then there exists $u(x,y)$ satisying $du = M(x,y)dx + N(x,y)dy$. Thus we have

$$M = u_{x}, N = u_{y}$$

Now partially differentiate $M$ with respect to $y$ and partially differentiate $N$ with respect to $x$. Then

$$M_{y} = u_{xy}, N_{x} = u_{yx}$$

Since $M_{y}, N_{x}$ are continuous, $u_{xy},u_{yx}$ are also continuous.Thus by Schwarz lemma, $u_{xy} = u_{yx}$ and $M_{y} = N_{x}$.
$(2) \Rightarrow (1)$
Let $(x_{0},y_{0})$ be a point in the domain of $M,N$. Consider

$$u(x,y) = \int_{x_0}^{x}M(x,y)dx + \int_{y_0}^{y}N(x_0,y)dy$$

Then $u_x(x,y) = M(x,y)$. Since $M_y = N_x$,

$$u_y(x,y) = \int_{x_0}^{x}M_y(x,y)dx + N(x_0,y) = \int_{x_0}^{x}N_x(x,y)dx + N(x_0,y)$$

$$= (N(x,y) - N(x_0,y)) + N(x_0,y) = N(x,y)$$

Therefore,

$$du = M(x,y)dx + N(x,y)dy$$

and $M(x,y)dx + N(x,y)dy = 0$ is exact differential equation. $\ \blacksquare$

In the proof above, the general solution of $u(x,y)$ is given.

$$u(x,y) = \int_{x_{0}}^{x}M(x,y)dx + \int_{y_{0}}^{y}N(x_{0},y)dy = c$$

Example 1..11   Find the general solution of the following differential equation

$$(y\cos{x} - \sin{y})dx + (\sin{x} - x\cos{y})dy = 0.$$

SOLUTION

$$M_{y} = \frac{\partial}{\partial y}(y\cos{x} - \sin{y}) = \cos{x} - \cos{y},$$

$$N_{x} = \frac{\partial}{\partial x}(\sin{x} - x\cos{y}) = \cos{x} - \cos{y}$$

Thus it is exact differential equation. Then set $(x_{0},y_{0}) = (0,0)$ to have

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

$$= \int_{0}^{x}(y\cos{x} - \sin{y})dx + \int_{0}^{y}0 dy$$

$$= y\sin{x} - x\sin{y}$$

Thus

$$y\sin{x} - x\sin{y} = c \ensuremath{\ \blacksquare}$$

Example 1..12   Solve the initial value problem.

$$(2xy - 3y)dx + (4y^{3} + x^{2} - 3x + 4)dy = 0, \ y(0) = 1.$$

SOLUTION Note that $M_{y} = 2x - 3 = N_{x}$,

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

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

$$= x^{2}y - 3xy + y^{4} + 4y .$$

Thus the general solution is

$$x^{2}y - 3xy + y^{4} + 4y = c$$

Since $y(0) = 1$, we have $c = 5$ $\ \blacksquare$

Instead of using the formula $u(x,y) = \int_{x_{0}}^{x}M(x,y)dx + \int_{y_{0}}^{y}N(x_{0},y)dy$, we introduce a simpler method called grouping method.

Example 1..13   Find the general solution of $(2x + 3y)dx + (3x + y^2 + 3)dy = 0$.

SOLUTION Note that $M_{y} = 3 = N_{x}$. Thus it is exact. Now we write $M(x,y),N(x,y)$ as

$$\underbrace{2x dx}_{x} + \underbrace{(3ydx + 3x dy)}_{x,y} + \underbrace{(y^2 + 3)dy}_{y} = 0$$

Now we write these by using the total differentiation.

$$d(x^2) + d(3xy) + d(\frac{y^3}{3} + 3y) = d(c) .$$

Then,

$$x^2 + 3xy + \frac{y^3}{3} + 3y = c \ensuremath{\ \blacksquare}$$