Exact differential equation

In mathematics, an exact differential equation or total differential equation is a certain kind of ordinary differential equation which is widely used in physics and engineering.

Definition

Given a simply connected and open subset D of R² and two functions I and J which are continuous on D, an implicit first-order ordinary differential equation of the form

${\displaystyle I(x,y)\,\mathrm {d} x+J(x,y)\,\mathrm {d} y=0,\,\!}$

is called an exact differential equation if there exists a continuously differentiable function F, called the potential function,[1][2] so that

${\displaystyle {\frac {\partial F}{\partial x}}=I}$

and

${\displaystyle {\frac {\partial F}{\partial y}}=J.}$

The nomenclature of "exact differential equation" refers to the exact differential of a function. For a function ${\displaystyle F(x_{0},x_{1},...,x_{n-1},x_{n})}$, the exact or total derivative with respect to ${\displaystyle x_{0}}$ is given by

${\displaystyle {\frac {\mathrm {d} F}{\mathrm {d} x_{0}}}={\frac {\partial F}{\partial x_{0}}}+\sum _{i=1}^{n}{\frac {\partial F}{\partial x_{i}}}{\frac {\mathrm {d} x_{i}}{\mathrm {d} x_{0}}}.}$

Example

The function ${\displaystyle F:\mathbb {R} ^{2}\to \mathbb {R} }$ given by

${\displaystyle F(x,y)={\frac {1}{2}}(x^{2}+y^{2})+c}$

is a potential function for the differential equation

${\displaystyle x\,\mathrm {d} x+y\,\mathrm {d} y=0.\,}$

Existence of potential functions

In physical applications the functions I and J are usually not only continuous but even continuously differentiable. Schwarz's Theorem then provides us with a necessary criterion for the existence of a potential function. For differential equations defined on simply connected sets the criterion is even sufficient and we get the following theorem:

Given a differential equation of the form (for example, when F has zero slope in the x and y direction at F(x,y) ):

${\displaystyle I(x,y)\,\mathrm {d} x+J(x,y)\,\mathrm {d} y=0,\,\!}$

with I and J continuously differentiable on a simply connected and open subset D of R² then a potential function F exists if and only if

${\displaystyle {\frac {\partial I}{\partial y}}(x,y)={\frac {\partial J}{\partial x}}(x,y).}$

Solutions to exact differential equations

Given an exact differential equation defined on some simply connected and open subset D of R² with potential function F, a differentiable function f with (x, f(x)) in D is a solution if and only if there exists real number c so that

${\displaystyle F(x,f(x))=c.\,}$

For an initial value problem

${\displaystyle y(x_{0})=y_{0}\,}$

we can locally find a potential function by

${\displaystyle F(x,y)=\int _{x_{0}}^{x}I(t,y_{0})\mathrm {d} t+\int _{y_{0}}^{y}J(x,t)\mathrm {d} t=\int _{x_{0}}^{x}I(t,y_{0})\mathrm {d} t+\int _{y_{0}}^{y}\left[J(x_{0},t)+\int _{x_{0}}^{x}{\frac {\partial I}{\partial t}}(u,t)\,\mathrm {d} u\,\right]\mathrm {d} t.}$

Solving

${\displaystyle F(x,y)=c\,}$

for y, where c is a real number, we can then construct all solutions.

Second order exact differential equations

The concept of exact differential equations can be extended to second order equations.[3] Consider starting with the first-order exact equation:

$${\displaystyle I(x,y)+J(x,y){dy \over dx}=0}$$

Since both functions ${\displaystyle I(x,y),J(x,y)}$ are functions of two variables, implicitly differentiating the multivariate function yields

$${\displaystyle {\operatorname {d} \!I \over \operatorname {d} \!x}+({\operatorname {d} \!J \over \operatorname {d} \!x}){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0}$$

Expanding the total derivatives gives that

$${\displaystyle {\operatorname {d} \!I \over \operatorname {d} \!x}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}}$$

and that

$${\displaystyle {\operatorname {d} \!J \over \operatorname {d} \!x}={\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx}}$$

Combining the ${\displaystyle {dy \over dx}}$ terms gives

$${\displaystyle {\partial I \over \partial x}+{dy \over dx}({\partial I \over \partial y}+{\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx})+{d^{2}y \over dx^{2}}(J(x,y))=0}$$

If the equation is exact, then ${\displaystyle {\partial J \over \partial x}={\partial I \over \partial y}}$ . Additionally, the total derivative of ${\displaystyle J(x,y)}$ is equal to its implicit ordinary derivative ${\displaystyle {dJ \over dx}}$. This leads to the rewritten equation

$${\displaystyle {\partial I \over \partial x}+{dy \over dx}({\partial J \over \partial x}+{dJ \over dx})+{d^{2}y \over dx^{2}}(J(x,y))=0}$$

Now, let there be some second-order differential equation

$${\displaystyle f(x,y)+g(x,y,{dy \over dx}){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0}$$

If ${\displaystyle {\partial J \over \partial x}={\partial I \over \partial y}}$ for exact differential equations, then

$${\displaystyle \int ({\partial I \over \partial y})dy=\int ({\partial J \over \partial x})dy}$$

and

$${\displaystyle \int ({\partial I \over \partial y})dy=\int ({\partial J \over \partial x})dy=I(x,y)-h(x)}$$

where ${\displaystyle h(x)}$ is some arbitrary function only of ${\displaystyle x}$ that was differentiated away to zero upon taking the partial derivative of ${\displaystyle I(x,y)}$ with respect to ${\displaystyle y}$. Although the sign on ${\displaystyle h(x)}$ could be positive, it is more intuitive to think of the integral's result as ${\displaystyle I(x,y)}$ that is missing some original extra function ${\displaystyle h(x)}$ that was partially differentiated to zero. Next, if

$${\displaystyle {\operatorname {d} \!I \over \operatorname {d} \!x}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}}$$

then the term ${\displaystyle {\partial I \over \partial x}}$ should be a function only of ${\displaystyle x}$ and ${\displaystyle y}$, since partial differentiation with respect to ${\displaystyle x}$ will hold ${\displaystyle y}$ constant and not produce any derivatives of ${\displaystyle y}$. In the second order equation

$${\displaystyle f(x,y)+g(x,y,{dy \over dx}){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0}$$

only the term ${\displaystyle f(x,y)}$ is a term purely of ${\displaystyle x}$ and ${\displaystyle y}$. Let ${\displaystyle {\partial I \over \partial x}=f(x,y)}$. If ${\displaystyle {\partial I \over \partial x}=f(x,y)}$, then

$${\displaystyle f(x,y)={\operatorname {d} \!I \over \operatorname {d} \!x}-{\partial I \over \partial y}{dy \over dx}}$$

Since the total derivative of ${\displaystyle I(x,y)}$ with respect to ${\displaystyle x}$ is equivalent to the implicit ordinary derivative ${\displaystyle {dI \over dx}}$ , then

$${\displaystyle f(x,y)+{\partial I \over \partial y}{dy \over dx}={dI \over dx}={d \over dx}(I(x,y)-h(x))+{dh(x) \over dx}}$$

So,

$${\displaystyle {dh(x) \over dx}=f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))}$$

and

$${\displaystyle h(x)=\int (f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x)))dx}$$

Thus, the second order differential equation

$${\displaystyle f(x,y)+g(x,y,{dy \over dx}){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0}$$

is exact only if ${\displaystyle g(x,y,{dy \over dx})={\operatorname {d} \!J \over \operatorname {d} \!x}+{\partial J \over \partial x}={dJ \over dx}+{\partial J \over \partial x}}$ and only if the below expression

$${\displaystyle \int (f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x)))dx=\int (f(x,y)-{\partial (I(x,y)-h(x)) \over \partial x})dx}$$

is a function solely of ${\displaystyle x}$. Once ${\displaystyle h(x)}$ is calculated with its arbitrary constant, it is added to ${\displaystyle I(x,y)-h(x)}$to make ${\displaystyle I(x,y)}$. If the equation is exact, then we can reduce to the first order exact form which is solvable by the usual method for first-order exact equations.

$${\displaystyle I(x,y)+J(x,y){dy \over dx}=0}$$

Now, however, in the final implicit solution there will be a ${\displaystyle C_{1}x}$ term from integration of ${\displaystyle h(x)}$ with respect to ${\displaystyle x}$ twice as well as a ${\displaystyle C_{2}}$, two arbitrary constants as expected from a second-order equation.

Example

Given the differential equation

$${\displaystyle (1-x^{2})y''-4xy'-2y=0}$$

one can always easily check for exactness by examining the ${\displaystyle y''}$ term. In this case, both the partial and total derivative of ${\displaystyle 1-x^{2}}$ with respect to ${\displaystyle x}$ are ${\displaystyle -2x}$, so their sum is ${\displaystyle -4x}$, which is exactly the term in front of ${\displaystyle y'}$. With one of the conditions for exactness met, one can calculate that

$${\displaystyle \int (-2x)dy=I(x,y)-h(x)=-2xy}$$

Letting ${\displaystyle f(x,y)=-2y}$, then

$${\displaystyle \int (-2y-2xy'-{d \over dx}(-2xy)dx=\int (-2y-2xy'+2xy'+2y)dx=\int (0)dx=h(x)}$$

So, ${\displaystyle h(x)}$ is indeed a function only of ${\displaystyle x}$ and the second order differential equation is exact. Therefore, ${\displaystyle h(x)=C_{1}}$ and ${\displaystyle I(x,y)=-2xy+C_{1}}$. Reduction to a first-order exact equation yields

$${\displaystyle -2xy+C_{1}+(1-x^{2})y'=0}$$

Integrating ${\displaystyle I(x,y)}$ with respect to ${\displaystyle x}$ yields

$${\displaystyle -x^{2}y+C_{1}x+i(y)=0}$$

where ${\displaystyle i(y)}$ is some arbitrary function of ${\displaystyle y}$. Differentiating with respect to ${\displaystyle y}$ gives an equation correlating the derivative and the ${\displaystyle y'}$ term.

$${\displaystyle -x^{2}+i'(y)=1-x^{2}}$$

So, ${\displaystyle i(y)=y+C_{2}}$ and the full implicit solution becomes

$${\displaystyle C_{1}x+C_{2}+y-x^{2}y=0}$$

Solving explicitly for ${\displaystyle y}$ yields

${\displaystyle y={\frac {C_{1}x+C_{2}}{1-x^{2}}}}$

Higher order exact differential equations

The concepts of exact differential equations can be extended to any order. Starting with the exact second order equation

$${\displaystyle {d^{2}y \over dx^{2}}(J(x,y))+{dy \over dx}({dJ \over dx}+{\partial J \over \partial x})+f(x,y)=0}$$

it was previously shown that equation is defined such that

$${\displaystyle f(x,y)={dh(x) \over dx}+{d \over dx}(I(x,y)-h(x))-{\partial J \over \partial x}{dy \over dx}}$$

Implicit differentiation of the exact second-order equation ${\displaystyle n}$ times will yield an ${\displaystyle (n+2)}$th order differential equation with new conditions for exactness that can be readily deduced from the form of the equation produced. For example, differentiating the above second-order differential equation once to yield a third-order exact equation gives the following form

$${\displaystyle {d^{3}y \over dx^{3}}(J(x,y))+{d^{2}y \over dx^{2}}{dJ \over dx}+{d^{2}y \over dx^{2}}({dJ \over dx}+{\partial J \over \partial x})+{dy \over dx}({d^{2}J \over dx^{2}}+{d \over dx}({\partial J \over \partial x}))+{df(x,y) \over dx}=0}$$

where

$${\displaystyle {df(x,y) \over dx}={d^{2}h(x) \over dx^{2}}+{d^{2} \over dx^{2}}(I(x,y)-h(x))-{d^{2}y \over dx^{2}}{\partial J \over \partial x}-{dy \over dx}{d \over dx}({\partial J \over \partial x})=F(x,y,{dy \over dx})}$$

and where ${\displaystyle F(x,y,{dy \over dx})}$is a function only of ${\displaystyle x,y}$ and ${\displaystyle {dy \over dx}}$. Combining all ${\displaystyle {dy \over dx}}$ and ${\displaystyle {d^{2}y \over dx^{2}}}$ terms not coming from ${\displaystyle F(x,y,{dy \over dx})}$ gives

$${\displaystyle {d^{3}y \over dx^{3}}(J(x,y))+{d^{2}y \over dx^{2}}(2{dJ \over dx}+{\partial J \over \partial x})+{dy \over dx}({d^{2}J \over dx^{2}}+{d \over dx}({\partial J \over \partial x}))+F(x,y,{dy \over dx})=0}$$

Thus, the three conditions for exactness for a third-order differential equation are: the ${\displaystyle {d^{2}y \over dx^{2}}}$ term must be ${\displaystyle 2{dJ \over dx}+{\partial J \over \partial x}}$, the ${\displaystyle {dy \over dx}}$ term must be ${\displaystyle {d^{2}J \over dx^{2}}+{d \over dx}({\partial J \over \partial x})}$ and

$${\displaystyle F(x,y,{dy \over dx})-{d^{2} \over dx^{2}}(I(x,y)-h(x))+{d^{2}y \over dx^{2}}{\partial J \over \partial x}+{dy \over dx}{d \over dx}({\partial J \over \partial x})}$$

must be a function solely of ${\displaystyle x}$.

Example

Consider the nonlinear third-order differential equation

$${\displaystyle yy'''+3y'y''+12x^{2}=0}$$

If ${\displaystyle J(x,y)=y}$, then ${\displaystyle y''(2{dJ \over dx}+{\partial J \over \partial x})}$ is ${\displaystyle 2y'y''}$ and ${\displaystyle y'({d^{2}J \over dx^{2}}+{d \over dx}({\partial J \over \partial x}))=y'y''}$which together sum to ${\displaystyle 3y'y''}$. Fortunately, this appears in our equation. For the last condition of exactness,

$${\displaystyle F(x,y,{dy \over dx})-{d^{2} \over dx^{2}}(I(x,y)-h(x))+{d^{2}y \over dx^{2}}{\partial J \over \partial x}+{dy \over dx}{d \over dx}({\partial J \over \partial x})=12x^{2}-0+0+0=12x^{2}}$$

which is indeed a function only of ${\displaystyle x}$. So, the differential equation is exact. Integrating twice yields that ${\displaystyle h(x)=x^{4}+C_{1}x+C_{2}=I(x,y)}$. Rewriting the equation as a first-order exact differential equation yields

$${\displaystyle x^{4}+C_{1}x+C_{2}+yy'=0}$$

Integrating ${\displaystyle I(x,y)}$ with respect to ${\displaystyle x}$ gives that ${\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+i(y)=0}$. Differentiating with respect to ${\displaystyle y}$ and equating that to the term in front of ${\displaystyle y'}$ in the first-order equation gives that

${\displaystyle i'(y)=y}$ and that ${\displaystyle i(y)={y^{2} \over 2}+C_{3}}$. The full implicit solution becomes

$${\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+C_{3}+{y^{2} \over 2}=0}$$

The explicit solution, then, is

$${\displaystyle y=\pm {\sqrt {C_{1}x^{2}+C_{2}x+C_{3}-{2x^{5} \over 5}}}}$$

See also

• Exact differential
• Inexact differential equation

References

1. Wolfgang Walter (11 March 2013). Ordinary Differential Equations. Springer Science & Business Media. ISBN 978-1-4612-0601-9.
2. Vladimir A. Dobrushkin (16 December 2014). Applied Differential Equations: The Primary Course. CRC Press. ISBN 978-1-4987-2835-5.
3. Tenenbaum, Morris; Pollard, Harry (1963). "Solution of the Linear Differential Equation with Nonconstant Coefficients. Reduction of Order Method.". Ordinary Differential Equations: An Elementary Textbook for Students of Mathematics, Engineering and the Sciences. New York: Dover. pp. 248. ISBN 0-486-64940-7.

Further reading

• Boyce, William E.; DiPrima, Richard C. (1986). Elementary Differential Equations (4th ed.). New York: John Wiley & Sons, Inc. ISBN 0-471-07894-8