Exact differential equations

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

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

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

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

Find

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

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

Integrate the first equation by $x$

to obtain

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

Note that

is an arbitrary function of $y$

. Now differentiate with respect ot $y$

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

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

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

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

. Therefore,

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

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

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

is the same as the differential $du$

of some function $u$

. Then the general solution is given by

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

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

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

Theorem 1..1 Let $M(x,y)$

and

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

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

Now partially differentiate $M$

with respect to $y$

and partially differentiate $N$

with respect to $x$

. Then

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

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

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

Then

. Since

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

Therefore,

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

and

is exact differential equation. $\ \blacksquare$

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

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

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

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

SOLUTION

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

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

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

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

Thus

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

Example 1..12 Solve the initial value problem.

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

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

$\displaystyle u(x,y)$	$\displaystyle =$	$\displaystyle \int_{0}^{x}M(x,y)dx + \int_{0}^{y}N(0,y)dy$
	$\displaystyle =$	$\displaystyle \int_{0}^{x}(2xy - 3y)dx + \int_{0}^{y}(4y^{3} + 4)dy$
	$\displaystyle =$	$\displaystyle x^{2}y - 3xy + y^{4} + 4y .$

Thus the general solution is

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

Since

, we have

$\ \blacksquare$

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

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

Now we write these by using the total differentiation.

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

Then,

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

Subsections

Exercise