Definition
Given a simply connected and open subset D of
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)\,dx+J(x,y)\,dy=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56b0331acfa2c34e8b43758c0004278a70310a26)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77182ff92c77806817b7d7114e4670c10b0a3b48)
and
![{\displaystyle {\frac {\partial F}{\partial y))=J.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a69187935a8440568611284d336bee365f5dbd3)
An exact equation may also be presented in the following form:
![{\displaystyle I(x,y)+J(x,y)\,y'(x)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8586e445082d3670e51ee05f4de56caecd5cdcf1)
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
, the exact or total derivative with respect to
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))}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4d8ef74c99793582573d90a0a0e061e5fac32a3)
Example
The function
given by
![{\displaystyle F(x,y)={\frac {1}{2))(x^{2}+y^{2})+c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a0f78b6d421426d3b89bc92afd28a684abe5ee8)
is a potential function for the differential equation
![{\displaystyle x\,dx+y\,dy=0.\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92fb9f68d0327136271814b556d9b15a2eae3c17)
First order exact differential equations
Identifying first order exact differential equations
Let the functions
,
,
, and
, where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region
. Then the differential equation
is exact if and only if
That is, there exists a function
, called a potential function, such that
So, in general:
Proof
The proof has two parts.
First, suppose there is a function
such that
It then follows that
Since
and
are continuous, then
and
are also continuous which guarantees their equality.
The second part of the proof involves the construction of
and can also be used as a procedure for solving first-order exact differential equations. Suppose that
and let there be a function
for which
Begin by integrating the first equation with respect to
. 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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f380ead829fb60bf9a6fe6f45241efa41b11518)
![{\displaystyle \psi (x,y)=\int {M(x,y)dx}+h(y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd0bc48651f17a87357009b81ceadcc02d8d8ccb)
![{\displaystyle \psi (x,y)=Q(x,y)+h(y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f24515f6013ce23f4a8a3c46b126d66164f6990c)
where
is any differentiable function such that
. The function
plays the role of a constant of integration, but instead of just a constant, it is function of
, since
is a function of both
and
and we are only integrating with respect to
.
Now to show that it is always possible to find an
such that
.
![{\displaystyle \psi (x,y)=Q(x,y)+h(y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f24515f6013ce23f4a8a3c46b126d66164f6990c)
Differentiate both sides with respect to
.
![{\displaystyle {\frac {\partial \psi }{\partial y))(x,y)={\frac {\partial Q}{\partial y))(x,y)+h'(y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/209500b0585cc34fc15855386f7183b46a9045e0)
Set the result equal to
and solve for
.
![{\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y))(x,y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0be8dc8e735a949056dabfc1378c0b36ca90b056)
In order to determine
from this equation, the right-hand side must depend only on
. This can be proven by showing that its derivative with respect to
is always zero, so differentiate the right-hand side with respect to
.
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34619153b5b62852157e0bdd2d20fe23e3ab3e23)
Since
,
![{\displaystyle {\frac {\partial N}{\partial x))(x,y)-{\frac {\partial M}{\partial y))(x,y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f9859d5952bce48a2e5ce08d48b7d936941ef5d)
Now, this is zero based on our initial supposition that
Therefore,
![{\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y))(x,y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0be8dc8e735a949056dabfc1378c0b36ca90b056)
![{\displaystyle h(y)=\int {\left(N(x,y)-{\frac {\partial Q}{\partial y))(x,y)\right)dy))](https://wikimedia.org/api/rest_v1/media/math/render/svg/48f7f4f5e0630b691eed4688b53727bf758f2ee8)
![{\displaystyle \psi (x,y)=Q(x,y)+\int {\left(N(x,y)-{\frac {\partial Q}{\partial y))(x,y)\right)dy}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8241f39f5b8f46ba1ea8c05e516b3b6555edfd1c)
And this completes the proof.
Solutions to first order exact differential equations
First order exact differential equations of the form
![{\displaystyle M(x,y)+N(x,y){\frac {dy}{dx))=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b0c2b9f42dc3a49861e9bb9e5053affb413f6eb)
can be written in terms of the potential function
![{\displaystyle {\frac {\partial \psi }{\partial x))+{\frac {\partial \psi }{\partial y)){\frac {dy}{dx))=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32e9df58e24a325056ee96871aeb889a1245bf70)
where
![{\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3bb263733c1306cbd81957396ab642bbcdcbd42)
This is equivalent to taking the exact differential of
.
![{\displaystyle {\frac {\partial \psi }{\partial x))+{\frac {\partial \psi }{\partial y)){\frac {dy}{dx))=0\iff {\frac {d}{dx))\psi (x,y(x))=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e76ef9a866f1e989c842895015a6d900010e645b)
The solutions to an exact differential equation are then given by
![{\displaystyle \psi (x,y(x))=c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/164dfa7db1d182b3199fb6d815dc29fb27d71123)
and the problem reduces to finding
.
This can be done by integrating the two expressions
and
and then writing down each term in the resulting expressions only once and summing them up in order to get
.
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))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3bb263733c1306cbd81957396ab642bbcdcbd42)
it follows, by integrating both sides, that
![{\displaystyle {\begin{cases}\psi (x,y)=\int {M(x,y)dx}+h(y)=Q(x,y)+h(y)\\\psi (x,y)=\int {N(x,y)dy}+g(x)=P(x,y)+g(x)\end{cases))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8eeb259782669938b98ad787785584c69a97ea14)
Therefore,
![{\displaystyle Q(x,y)+h(y)=P(x,y)+g(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10c7e64f5df9a12231c747d54393d8f1d08c2b09)
where
and
are differentiable functions such that
and
.
In order for this to be true and for both sides to result in the exact same expression, namely
, then
must be contained within the expression for
because it cannot be contained within
, since it is entirely a function of
and not
and is therefore not allowed to have anything to do with
. By analogy,
must be contained within the expression
.
Ergo,
![{\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and ))P(x,y)=h(y)+d(x,y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30cae2be470464f6ba039cb44f8d41630cbea387)
for some expressions
and
.
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/494544962d34abff54cca349eb9bd0dd2f77193a)
and so
and
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30483d0b936b6dc317ce6b8779522017a1277661)
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))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23bfe0ccfb7a727eef5b4cb434db0353c19ad120)
it follows that
![{\displaystyle \psi (x,y)=g(x)+f(x,y)+h(y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fba0397bf3e276c9cb69e50cd1d92f6767122d28)
So, we can construct
by doing
and
and then taking the common terms we find within the two resulting expressions (that would be
) and then adding the terms which are uniquely found in either one of them -
and
.
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\left(x,y\right)+J\left(x,y\right){dy \over dx}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b3fa17eaeb00f5258a52ee9de6f7f1649c9412b)
Since both functions
,
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))\left(J\left(x,y\right)\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1086db92cf066a991ef96166752a126eac09a9bb)
Expanding the total derivatives gives that
![{\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx))](https://wikimedia.org/api/rest_v1/media/math/render/svg/24056990358ad975915f62b4af74958e8f00c6e4)
and that
![{\displaystyle {dJ \over dx}={\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx))](https://wikimedia.org/api/rest_v1/media/math/render/svg/29d5225eb858c33e3d42898b915d597e7ccbd5e4)
Combining the
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))\left(J\left(x,y\right)\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee7e6ecf1663a1b70cee46e96c33d9cd96bf68b5)
If the equation is exact, then
. Additionally, the total derivative of
is equal to its implicit ordinary derivative
. 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))\left(J\left(x,y\right)\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4848827819f8f26652be7b9a5982be6115b5a9c)
Now, let there be some second-order differential equation
![{\displaystyle f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9022d3ca57ff406fe0111be119c21289619eebda)
If
for exact differential equations, then
![{\displaystyle \int \left({\partial I \over \partial y}\right)dy=\int \left({\partial J \over \partial x}\right)dy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8248c659fbe6db8d1e7aa43dfd632ab375ccad9)
and
![{\displaystyle \int \left({\partial I \over \partial y}\right)dy=\int \left({\partial J \over \partial x}\right)dy=I\left(x,y\right)-h\left(x\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39540195893062eaf74c25f7d693ebca098a2185)
where
is some arbitrary function only of
that was differentiated away to zero upon taking the partial derivative of
with respect to
. Although the sign on
could be positive, it is more intuitive to think of the integral's result as
that is missing some original extra function
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))](https://wikimedia.org/api/rest_v1/media/math/render/svg/24056990358ad975915f62b4af74958e8f00c6e4)
then the term
should be a function only of
and
, since partial differentiation with respect to
will hold
constant and not produce any derivatives of
. In the second order equation
![{\displaystyle f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9022d3ca57ff406fe0111be119c21289619eebda)
only the term
is a term purely of
and
. Let
. If
, then
![{\displaystyle f\left(x,y\right)={dI \over dx}-{\partial I \over \partial y}{dy \over dx))](https://wikimedia.org/api/rest_v1/media/math/render/svg/aec979a672b595bc641aacdf5209329101748128)
Since the total derivative of
with respect to
is equivalent to the implicit ordinary derivative
, then
![{\displaystyle f\left(x,y\right)+{\partial I \over \partial y}{dy \over dx}={dI \over dx}={d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)+{dh\left(x\right) \over dx))](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c3a5afca52790dd524f0e1e27dfc32d805e7bbf)
So,
![{\displaystyle {dh\left(x\right) \over dx}=f\left(x,y\right)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb1240d37ee264d8a2206be03a66e0e8e293f2fc)
and
![{\displaystyle h\left(x\right)=\int \left(f\left(x,y\right)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fe49e96a2445dcb5c042eec5b9f8ad56d0c5e75)
Thus, the second order differential equation
![{\displaystyle f\left(x,y\right)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2))\left(J\left(x,y\right)\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9022d3ca57ff406fe0111be119c21289619eebda)
is exact only if
and only if the below expression
![{\displaystyle \int \left(f\left(x,y\right)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx=\int \left(f\left(x,y\right)-{\partial \left(I\left(x,y\right)-h\left(x\right)\right) \over \partial x}\right)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81b5999c0fd7f49822addbaf472c79136205a566)
is a function solely of
. Once
is calculated with its arbitrary constant, it is added to
to make
. 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\left(x,y\right)+J\left(x,y\right){dy \over dx}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b3fa17eaeb00f5258a52ee9de6f7f1649c9412b)
Now, however, in the final implicit solution there will be a
term from integration of
with respect to
twice as well as a
, two arbitrary constants as expected from a second-order equation.
Example
Given the differential equation
![{\displaystyle \left(1-x^{2}\right)y''-4xy'-2y=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e730af5b7bc9a4df8bcc533aaf1655ab46364304)
one can always easily check for exactness by examining the
term. In this case, both the partial and total derivative of
with respect to
are
, so their sum is
, which is exactly the term in front of
. With one of the conditions for exactness met, one can calculate that
![{\displaystyle \int \left(-2x\right)dy=I\left(x,y\right)-h\left(x\right)=-2xy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2ba35dc573eaaf6c9f6bcfb6e308a61f99e0cad)
Letting
, then
![{\displaystyle \int \left(-2y-2xy'-{d \over dx}\left(-2xy\right)\right)dx=\int \left(-2y-2xy'+2xy'+2y\right)dx=\int \left(0\right)dx=h\left(x\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2429644ee3015f47af7c18243ad5ce0cab9446a)
So,
is indeed a function only of
and the second order differential equation is exact. Therefore,
and
. Reduction to a first-order exact equation yields
![{\displaystyle -2xy+C_{1}+\left(1-x^{2}\right)y'=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0362e1f09b29d037ada8c3523cdd0b6b020a45b1)
Integrating
with respect to
yields
![{\displaystyle -x^{2}y+C_{1}x+i\left(y\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a97c6c72d183336921babb4fba4e5262d8f06211)
where
is some arbitrary function of
. Differentiating with respect to
gives an equation correlating the derivative and the
term.
![{\displaystyle -x^{2}+i'\left(y\right)=1-x^{2))](https://wikimedia.org/api/rest_v1/media/math/render/svg/c26b0201e35bdd540fb2cdb4fefc9cf3b21ea9eb)
So,
and the full implicit solution becomes
![{\displaystyle C_{1}x+C_{2}+y-x^{2}y=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c364b9816cbfb5f6129535a6948f7cfe8132e30b)
Solving explicitly for
yields
![{\displaystyle y={\frac {C_{1}x+C_{2)){1-x^{2))))](https://wikimedia.org/api/rest_v1/media/math/render/svg/87d9cabde4e2055be75e8e4b28b7f932a6f12645)
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))\left(J\left(x,y\right)\right)+{dy \over dx}\left({dJ \over dx}+{\partial J \over \partial x}\right)+f\left(x,y\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97f65ecf18969baa13abbe6e1d2e30742b6804e2)
it was previously shown that equation is defined such that
![{\displaystyle f\left(x,y\right)={dh\left(x\right) \over dx}+{d \over dx}\left(I\left(x,y\right)-h\left(x\right)\right)-{\partial J \over \partial x}{dy \over dx))](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4222a7feeb9eb570ceb6e3ef1169f3127ebc6b6)
Implicit differentiation of the exact second-order equation
times will yield an
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))\left(J\left(x,y\right)\right)+{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\left(x,y\right) \over dx}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/400e6a391cbe7f8955cd2f022540aa55f9561bbd)
where
![{\displaystyle {df\left(x,y\right) \over dx}={d^{2}h\left(x\right) \over dx^{2))+{d^{2} \over dx^{2))\left(I\left(x,y\right)-h\left(x\right)\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)=F\left(x,y,{dy \over dx}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04c5758b445a42bcacd9b20a8a6372a757c4e6fa)
and where
is a function only of
and
. Combining all
and
terms not coming from
gives
![{\displaystyle {d^{3}y \over dx^{3))\left(J\left(x,y\right)\right)+{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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80dc129d45de899265b76e1da185c48e3a67ad89)
Thus, the three conditions for exactness for a third-order differential equation are: the
term must be
, the
term must be
and
![{\displaystyle F\left(x,y,{dy \over dx}\right)-{d^{2} \over dx^{2))\left(I\left(x,y\right)-h\left(x\right)\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcda16e9bb41312045f6dd73f9e7721e64c38a4b)
must be a function solely of
.
Example
Consider the nonlinear third-order differential equation
![{\displaystyle yy'''+3y'y''+12x^{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24ff9bb3096af999f0559cd6435430da37865f8e)
If
, then
is
and
which together sum to
. 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\left(x,y\right)-h\left(x\right)\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))](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9325f813d3fa0b14e0051eceb4e5ce841fe4f1f)
which is indeed a function only of
. So, the differential equation is exact. Integrating twice yields that
. Rewriting the equation as a first-order exact differential equation yields
![{\displaystyle x^{4}+C_{1}x+C_{2}+yy'=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f1a8546245effb99e8e6ff32fc03a10042745f7)
Integrating
with respect to
gives that
. Differentiating with respect to
and equating that to the term in front of
in the first-order equation gives that
and that
. The full implicit solution becomes
![{\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+C_{3}+{y^{2} \over 2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8321aac32851151ac5fffcd6ed10a96cfbcc9d84)
The explicit solution, then, is
![{\displaystyle y=\pm {\sqrt {C_{1}x^{2}+C_{2}x+C_{3}-{\frac {2x^{5)){5))))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a4df0091528a2ce3653f59aaf6d4770e2ccffe8)