Exact differential equation

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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.

Given a simply connected and open subset D of ${\displaystyle \mathbb {R} ^{2}}${\displaystyle \mathbb {R} ^{2}} and two functions I and J which are continuous on D, an implicit first-order ordinary differential equation of the form

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

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

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

and

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

An exact equation may also be presented in the following form:

${\displaystyle I(x,y)+J(x,y),円y'(x)=0}${\displaystyle I(x,y)+J(x,y),円y'(x)=0}

where the same constraints on I and J apply for the differential equation to be exact.

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

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

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

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

is a potential function for the differential equation

${\displaystyle x,円dx+y,円dy=0.,円}${\displaystyle x,円dx+y,円dy=0.,円}

Let the functions ${\textstyle M}${\textstyle M}, ${\textstyle N}${\textstyle N}, ${\textstyle M_{y}}${\textstyle M_{y}}, and ${\textstyle N_{x}}${\textstyle N_{x}}, where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region ${\textstyle R:\alpha <x<\beta ,\gamma <y<\delta }${\textstyle R:\alpha <x<\beta ,\gamma <y<\delta }. Then the differential equation

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

is exact if and only if

${\displaystyle M_{y}(x,y)=N_{x}(x,y)}${\displaystyle M_{y}(x,y)=N_{x}(x,y)}

That is, there exists a function ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)}, called a potential function, such that

${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}

So, in general:

${\displaystyle M_{y}(x,y)=N_{x}(x,y)\iff {\begin{cases}\exists \psi (x,y)\\\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}${\displaystyle M_{y}(x,y)=N_{x}(x,y)\iff {\begin{cases}\exists \psi (x,y)\\\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}

The proof has two parts.

First, suppose there is a function ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)} such that ${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}

It then follows that ${\displaystyle M_{y}(x,y)=\psi _{xy}(x,y){\text{ and }}N_{x}(x,y)=\psi _{yx}(x,y)}${\displaystyle M_{y}(x,y)=\psi _{xy}(x,y){\text{ and }}N_{x}(x,y)=\psi _{yx}(x,y)}

Since ${\displaystyle M_{y}}${\displaystyle M_{y}} and ${\displaystyle N_{x}}${\displaystyle N_{x}} are continuous, then ${\displaystyle \psi _{xy}}${\displaystyle \psi _{xy}} and ${\displaystyle \psi _{yx}}${\displaystyle \psi _{yx}} are also continuous which guarantees their equality.

The second part of the proof involves the construction of ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)} and can also be used as a procedure for solving first-order exact differential equations. Suppose that ${\displaystyle M_{y}(x,y)=N_{x}(x,y)}${\displaystyle M_{y}(x,y)=N_{x}(x,y)} and let there be a function ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)} for which ${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}${\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}

Begin by integrating the first equation with respect to ${\displaystyle x}${\displaystyle x}. In practice, it doesn't matter if you integrate the first or the second equation, so long as the integration is done with respect to the appropriate variable.

$${\displaystyle {\frac {\partial \psi }{\partial x}}(x,y)=M(x,y)}$${\displaystyle {\frac {\partial \psi }{\partial x}}(x,y)=M(x,y)} $${\displaystyle \psi (x,y)=\int M(x,y),円dx+h(y)}$${\displaystyle \psi (x,y)=\int M(x,y),円dx+h(y)} $${\displaystyle \psi (x,y)=Q(x,y)+h(y)}$${\displaystyle \psi (x,y)=Q(x,y)+h(y)}

where ${\displaystyle Q(x,y)}${\displaystyle Q(x,y)} is any differentiable function such that ${\displaystyle Q_{x}=M}${\displaystyle Q_{x}=M}. The function ${\displaystyle h(y)}${\displaystyle h(y)} plays the role of a constant of integration, but instead of just a constant, it is function of ${\displaystyle y}${\displaystyle y}, since ${\displaystyle M}${\displaystyle M} is a function of both ${\displaystyle x}${\displaystyle x} and ${\displaystyle y}${\displaystyle y} and we are only integrating with respect to ${\displaystyle x}${\displaystyle x}.

Now to show that it is always possible to find an ${\displaystyle h(y)}${\displaystyle h(y)} such that ${\displaystyle \psi _{y}=N}${\displaystyle \psi _{y}=N}. $${\displaystyle \psi (x,y)=Q(x,y)+h(y)}$${\displaystyle \psi (x,y)=Q(x,y)+h(y)}

Differentiate both sides with respect to ${\displaystyle y}${\displaystyle y}. $${\displaystyle {\frac {\partial \psi }{\partial y}}(x,y)={\frac {\partial Q}{\partial y}}(x,y)+h'(y)}$${\displaystyle {\frac {\partial \psi }{\partial y}}(x,y)={\frac {\partial Q}{\partial y}}(x,y)+h'(y)}

Set the result equal to ${\displaystyle N}${\displaystyle N} and solve for ${\displaystyle h'(y)}${\displaystyle h'(y)}. $${\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)}$${\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)}

In order to determine ${\displaystyle h'(y)}${\displaystyle h'(y)} from this equation, the right-hand side must depend only on ${\displaystyle y}${\displaystyle y}. This can be proven by showing that its derivative with respect to ${\displaystyle x}${\displaystyle x} is always zero, so differentiate the right-hand side with respect to ${\displaystyle x}${\displaystyle x}. $${\displaystyle {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial }{\partial x}}{\frac {\partial Q}{\partial y}}(x,y)\iff {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial }{\partial y}}{\frac {\partial Q}{\partial x}}(x,y)}$${\displaystyle {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial }{\partial x}}{\frac {\partial Q}{\partial y}}(x,y)\iff {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial }{\partial y}}{\frac {\partial Q}{\partial x}}(x,y)}

Since ${\displaystyle Q_{x}=M}${\displaystyle Q_{x}=M}, $${\displaystyle {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial M}{\partial y}}(x,y)}$${\displaystyle {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial M}{\partial y}}(x,y)} Now, this is zero based on our initial supposition that ${\displaystyle M_{y}(x,y)=N_{x}(x,y)}${\displaystyle M_{y}(x,y)=N_{x}(x,y)}

Therefore, $${\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)}$${\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)} $${\displaystyle h(y)=\int {\left(N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)\right)dy}}$${\displaystyle h(y)=\int {\left(N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)\right)dy}}

$${\displaystyle \psi (x,y)=Q(x,y)+\int \left(N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)\right),円dy+C}$${\displaystyle \psi (x,y)=Q(x,y)+\int \left(N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)\right),円dy+C}

And this completes the proof.

First-order exact differential equations of the form $${\displaystyle M(x,y)+N(x,y){\frac {dy}{dx}}=0}$${\displaystyle M(x,y)+N(x,y){\frac {dy}{dx}}=0}

can be written in terms of the potential function ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)} $${\displaystyle {\frac {\partial \psi }{\partial x}}+{\frac {\partial \psi }{\partial y}}{\frac {dy}{dx}}=0}$${\displaystyle {\frac {\partial \psi }{\partial x}}+{\frac {\partial \psi }{\partial y}}{\frac {dy}{dx}}=0}

where $${\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}$${\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}

This is equivalent to taking the total derivative of ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)}. $${\displaystyle {\frac {\partial \psi }{\partial x}}+{\frac {\partial \psi }{\partial y}}{\frac {dy}{dx}}=0\iff {\frac {d}{dx}}\psi (x,y(x))=0}$${\displaystyle {\frac {\partial \psi }{\partial x}}+{\frac {\partial \psi }{\partial y}}{\frac {dy}{dx}}=0\iff {\frac {d}{dx}}\psi (x,y(x))=0}

The solutions to an exact differential equation are then given by $${\displaystyle \psi (x,y(x))=c}$${\displaystyle \psi (x,y(x))=c}

and the problem reduces to finding ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)}.

This can be done by integrating the two expressions ${\displaystyle M(x,y),円dx}${\displaystyle M(x,y),円dx} and ${\displaystyle N(x,y),円dy}${\displaystyle N(x,y),円dy} and then writing down each term in the resulting expressions only once and summing them up in order to get ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)}.

The reasoning behind this is the following. Since $${\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}$${\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}

it follows, by integrating both sides, that $${\displaystyle {\begin{cases}\psi (x,y)=\int M(x,y),円dx+h(y)=Q(x,y)+h(y)\\\psi (x,y)=\int N(x,y),円dy+g(x)=P(x,y)+g(x)\end{cases}}}$${\displaystyle {\begin{cases}\psi (x,y)=\int M(x,y),円dx+h(y)=Q(x,y)+h(y)\\\psi (x,y)=\int N(x,y),円dy+g(x)=P(x,y)+g(x)\end{cases}}}

Therefore, $${\displaystyle Q(x,y)+h(y)=P(x,y)+g(x)}$${\displaystyle Q(x,y)+h(y)=P(x,y)+g(x)}

where ${\displaystyle Q(x,y)}${\displaystyle Q(x,y)} and ${\displaystyle P(x,y)}${\displaystyle P(x,y)} are differentiable functions such that ${\displaystyle Q_{x}=M}${\displaystyle Q_{x}=M} and ${\displaystyle P_{y}=N}${\displaystyle P_{y}=N}.

In order for this to be true and for both sides to result in the exact same expression, namely ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)}, then ${\displaystyle h(y)}${\displaystyle h(y)} must be contained within the expression for ${\displaystyle P(x,y)}${\displaystyle P(x,y)} because it cannot be contained within ${\displaystyle g(x)}${\displaystyle g(x)}, since it is entirely a function of ${\displaystyle y}${\displaystyle y} and not ${\displaystyle x}${\displaystyle x} and is therefore not allowed to have anything to do with ${\displaystyle x}${\displaystyle x}. By analogy, ${\displaystyle g(x)}${\displaystyle g(x)} must be contained within the expression ${\displaystyle Q(x,y)}${\displaystyle Q(x,y)}.

Ergo, $${\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and }}P(x,y)=h(y)+d(x,y)}$${\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and }}P(x,y)=h(y)+d(x,y)}

for some expressions ${\displaystyle f(x,y)}${\displaystyle f(x,y)} and ${\displaystyle d(x,y)}${\displaystyle d(x,y)}. Plugging in into the above equation, we find that $${\displaystyle g(x)+f(x,y)+h(y)=h(y)+d(x,y)+g(x)\Rightarrow f(x,y)=d(x,y)}$${\displaystyle g(x)+f(x,y)+h(y)=h(y)+d(x,y)+g(x)\Rightarrow f(x,y)=d(x,y)} and so ${\displaystyle f(x,y)}${\displaystyle f(x,y)} and ${\displaystyle d(x,y)}${\displaystyle d(x,y)} turn out to be the same function. Therefore, $${\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and }}P(x,y)=h(y)+f(x,y)}$${\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and }}P(x,y)=h(y)+f(x,y)}

Since we already showed that $${\displaystyle {\begin{cases}\psi (x,y)=Q(x,y)+h(y)\\\psi (x,y)=P(x,y)+g(x)\end{cases}}}$${\displaystyle {\begin{cases}\psi (x,y)=Q(x,y)+h(y)\\\psi (x,y)=P(x,y)+g(x)\end{cases}}}

it follows that $${\displaystyle \psi (x,y)=g(x)+f(x,y)+h(y)}$${\displaystyle \psi (x,y)=g(x)+f(x,y)+h(y)}

So, we can construct ${\displaystyle \psi (x,y)}${\displaystyle \psi (x,y)} by doing ${\displaystyle \int M(x,y),円dx}${\displaystyle \int M(x,y),円dx} and ${\displaystyle \int N(x,y),円dy}${\displaystyle \int N(x,y),円dy} and then taking the common terms we find within the two resulting expressions (that would be ${\displaystyle f(x,y)}${\displaystyle f(x,y)} ) and then adding the terms which are uniquely found in either one of them – ${\displaystyle g(x)}${\displaystyle g(x)} and ${\displaystyle h(y)}${\displaystyle h(y)}.

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

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

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

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

Expanding the total derivatives gives that

${\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}}${\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}}

and that

${\displaystyle {dJ \over dx}={\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx}}${\displaystyle {dJ \over dx}={\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx}}

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

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

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

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

Now, let there be some second-order differential equation

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

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

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

and

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

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

Next, if

${\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}}${\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}}

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

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

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

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

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

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

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

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

Thus, the second-order differential equation

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

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

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

is a function solely of ${\displaystyle x}${\displaystyle x}. Once ${\displaystyle h(x)}${\displaystyle h(x)} is calculated with its arbitrary constant, it is added to ${\displaystyle I(x,y)-h(x)}${\displaystyle I(x,y)-h(x)} to make ${\displaystyle I(x,y)}${\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.

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

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

Given the differential equation

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

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

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

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

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

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

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

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

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

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

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

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

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

Solving explicitly for ${\displaystyle y}${\displaystyle y} yields

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

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

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

it was previously shown that equation is defined such that

${\displaystyle f(x,yt)={dht(x) \over dx}+{d \over dx}(I(x,y)-h(x))-{\partial J \over \partial x}{dy \over dx}}${\displaystyle f(x,yt)={dht(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 ${\displaystyle n}${\displaystyle n} times will yield an ${\displaystyle (n+2)}${\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

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

where

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

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

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

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

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

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

Consider the nonlinear third-order differential equation

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

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

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

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

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

Integrating ${\displaystyle I(x,y)}${\displaystyle I(x,y)} with respect to ${\displaystyle x}${\displaystyle x} gives that ${\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+i(y)=0}${\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+i(y)=0}. Differentiating with respect to ${\displaystyle y}${\displaystyle y} and equating that to the term in front of ${\displaystyle y'}${\displaystyle y'} in the first-order equation gives that ${\displaystyle i'(y)=y}${\displaystyle i'(y)=y} and that ${\displaystyle i(y)={y^{2} \over 2}+C_{3}}${\displaystyle i(y)={y^{2} \over 2}+C_{3}}. The full implicit solution becomes

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

The explicit solution, then, is

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

• Exact differential
• Inexact differential equation

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

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