Exact differential equation

First-order exact differential equations

Identifying first-order exact differential equations

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

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

is exact if and only if

${\displaystyle M_y(x,y)=N_x(x,y)}$

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

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

So, in general:

${\displaystyle M_y(x,y)=N_x(x,y)⟺\{\begin{array}{l}\mathrm{\exists }\psi (x,y)\\ {\psi }_x(x,y)=M(x,y)\\ {\psi }_y(x,y)=N(x,y)\end{array}}$

Proof

The proof has two parts.

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

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

Since ${\displaystyle M_y}$ and ${\displaystyle N_x}$ are continuous, then ${\displaystyle {\psi }_{xy}}$ and ${\displaystyle {\psi }_{yx}}$ are also continuous which guarantees their equality.

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

Begin by integrating the first equation with respect to ${\displaystyle x}$. 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.

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

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

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

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

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

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

Since ${\displaystyle Q_x=M}$, $${\displaystyle \frac{\mathrm{\partial }N}{\mathrm{\partial }x}(x,y)-\frac{\mathrm{\partial }M}{\mathrm{\partial }y}(x,y)}$$ Now, this is zero based on our initial supposition that ${\displaystyle M_y(x,y)=N_x(x,y)}$

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

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

And this completes the proof.

Solutions to first-order exact differential equations

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

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

where $${\displaystyle \{\begin{array}{l}{\psi }_x(x,y)=M(x,y)\\ {\psi }_y(x,y)=N(x,y)\end{array}}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Second-order exact differential equations

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

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

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

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

Expanding the total derivatives gives that

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

and that

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

Combining the $\frac{dy}{dx}$ terms gives

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

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

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

Now, let there be some second-order differential equation

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

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

${\displaystyle \int \left(\frac{\mathrm{\partial }I}{\mathrm{\partial }y}\right)dy=\int \left(\frac{\mathrm{\partial }J}{\mathrm{\partial }x}\right)dy}$

and

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

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

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

then the term ${\displaystyle \frac{\mathrm{\partial }I}{\mathrm{\partial }x}}$ 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

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

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

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

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

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

So,

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

and

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

Thus, the second-order differential equation

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

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

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

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.

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

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

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

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 ${\displaystyle y^{\prime }}$. 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

${\displaystyle \int \left(-2y-2x{y}^{\prime }-\frac{d}{dx}(-2xy)\right)dx=\int (-2y-2x{y}^{\prime }+2x{y}^{\prime }+2y)dx=\int (0)dx=h(x)}$

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

${\displaystyle -2xy+C_1+(1-x^2)y^{\prime }=0}$

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 ${\displaystyle y^{\prime }}$ term.

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

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

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

it was previously shown that equation is defined such that

${\displaystyle f(x,yt)=\frac{dht(x)}{dx}+\frac{d}{dx}(I(x,y)-h(x))-\frac{\mathrm{\partial }J}{\mathrm{\partial }x}\frac{dy}{dx}}$

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

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

where

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

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

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

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

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

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

Example

Consider the nonlinear third-order differential equation

${\displaystyle y{y}^{‴}+3y^{\prime }{y}^{″}+12x^2=0}$

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

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

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

${\displaystyle x^4+C_{1}x+C_2+y{y}^{\prime }=0}$

Integrating ${\displaystyle I(x,y)}$ with respect to ${\displaystyle x}$ gives that ${\displaystyle \frac{x^5}{5}+C_1{x}^2+C_{2}x+i(y)=0}$. Differentiating with respect to ${\displaystyle y}$ and equating that to the term in front of ${\displaystyle y^{\prime }}$ in the first-order equation gives that ${\displaystyle i^{\prime }(y)=y}$ and that ${\displaystyle i(y)=\frac{y^2}{2}+C_3}$. The full implicit solution becomes

${\displaystyle \frac{x^5}{5}+C_1{x}^2+C_{2}x+C_3+\frac{y^2}{2}=0}$

The explicit solution, then, is

${\displaystyle y=\pm \sqrt{C_1{x}^2+C_{2}x+C_3-\frac{2x^5}{5}}}$