How to find the solution to an exact differential equation
Exact differential equations have a specific format, and are solved using a specific set of steps
In order for a differential equation to be called an exact differential equation, it must be given in the form
???M(x,y)+N(x,y)\frac{dy}{dx}=0???
and have a solution ???\Psi(x,y)??? such that
???\frac{\partial\Psi}{\partial x}=\Psi_x=M(x,y)=M???
and
???\frac{\partial\Psi}{\partial y}=\Psi_y=N(x,y)=N???
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If we assume that ???\Psi(x,y)??? is a continuous function, then we can also assume that its mixed second-order partial derivatives are equal.
???\frac{\partial^2\Psi}{\partial x\partial y}=\frac{\partial^2\Psi}{\partial y\partial x}???
We can rewrite this equation in a different form as
???\Psi_{xy}=\Psi_{yx}???
???\left(\Psi_x\right)_y=\left(\Psi_y\right)_x???
???(M)_y=(N)_x???
???M_y=N_x???
In other words, we just used the fact that ???\Psi(x,y)??? has to be a solution to the exact differential equation to develop a test for exact differential equations, and the test tells us that a differential equation is exact if ???M_y=N_x???. So, if the partial derivative of ???M??? with respect to ???y??? is equal to the partial derivative of ???N??? with respect to ???x???, then the differential equation is exact.
This is a test we can use to say whether or not the differential equation is exact, before we go about finding the solution ???\Psi(x,y)???.
The general or implicit solution to an exact differential equation is given by
???\Psi(x,y)=c???
where ???c??? is a constant. If we want to, we can prove that this is the solution by starting with the standard form of an exact differential equation
???M(x,y)+N(x,y)\frac{dy}{dx}=0???
We already said that
???\frac{\partial\Psi}{\partial x}=\Psi_x=M(x,y)=M???
and
???\frac{\partial\Psi}{\partial y}=\Psi_y=N(x,y)=N???
so we can substitute into the standard form and get
???\Psi_x+\Psi_y\frac{dy}{dx}=0???
Using chain rule for multivariable functions, we can change the left-hand side to
???\frac{d}{dx}\left[\Psi(x,y)\right]=0???
Integrating both sides with respect to ???x??? to get ???\Psi(x,y)??? by itself, we get
???\int\frac{d}{dx}\left[\Psi(x,y)\right]\ dx=\int 0\ dx???
???\Psi(x,y)+c_1=c_2???
???\Psi(x,y)=c_2-c_1???
???\Psi(x,y)=c???
Now that we’ve proven that ???\Psi(x,y)=c??? is always the solution to an exact differential equation, we need to work on finding ???\Psi(x,y)???.
We’ve already used the fact that
???\frac{\partial\Psi}{\partial x}=\Psi_x=M(x,y)=M???
and
???\frac{\partial\Psi}{\partial y}=\Psi_y=N(x,y)=N???
and we’ll use it again here to find the solution. We’ll start with
???\Psi_x=M(x,y)???
and
???\Psi_y=N(x,y)???
The left-hand sides of both of these are partial derivatives of ???\Psi???, but we need to get back to ???\Psi??? itself. To do that, we’d need to take the integrals
???\int\Psi_x\ dx=\int M(x,y)\ dx???
and
???\int\Psi_y\ dy=\int N(x,y)\ dy???
Taking the integral with respect to ???x??? of the partial derivative with respect to ???x??? cancels both operations and leaves us with just ???\Psi???, in the same way that taking the integral with respect to ???y??? of the partial derivative with respect to ???y??? cancels those operations and leaves us with just ???\Psi???.
???\Psi=\int M(x,y)\ dx???
and
???\Psi=\int N(x,y)\ dy???
In other words, if we want to find an equation for ???\Psi???, we can either take the integral of ???M??? with respect to ???x???, or the integral of ???N??? with respect to ???y???. Both integrals will work, so we should look at ???M??? and ???N??? and then choose whichever function will be easier to integrate.
If we use the first integral, the one with ???M???, we have to remember that we’re integrating a multivariable function in terms of ???x??? and ???y??? with respect to ???x??? only. Which means that, instead of adding ???+C??? to account for the constant of integration after we integrate, we’ll have to add ???+h(y)??? to account for a function in terms of ???y???.
Similarly, if we use the second integral, the one with ???N???, we have to remember that we’re integrating a multivariable function in terms of ???x??? and ???y??? with respect to ???y??? only. Which means that, instead of adding ???+C??? to account for the constant of integration after we integrate, we’ll have to add ???+h(x)??? to account for a function in terms of ???x???.
Then it’s just a matter of solving for ???h(y)??? or ???h(x)???, which we’ll do by differentiating the equation for ???\Psi??? with respect to ???y??? if we’re trying to find ???h(y)???, or with respect to ???x??? if we’re trying to find ???h(x)???.
That differentiation process will give us either ???\Psi_x??? with ???h\prime(x)??? or ???\Psi_y??? with ???h\prime(y)???. We’ll substitute ???M(x,y)??? for ???\Psi_x??? or ???N(x,y)??? for ???\Psi_y???, and then simplify the equation to solve for ???h\prime(y)??? or ???h\prime(x)???. Then we can integrate both sides of the remaining equation to solve for ???h(y)??? or ???h(x)???.
Finally, we’ll plug ???h(y)??? or ???h(x)??? back into the equation for ???\Psi???, set the equation equal to ???c???, and this will be the general or implicit solution to the exact differential equation.
In summary, to find the solution to an exact differential equation, we’ll
Verify that ???M_y=N_x??? to confirm the differential equation is exact.
Use ???\Psi=\int M(x,y)\ dx??? or ???\Psi=\int N(x,y)\ dy??? to find ???\Psi(x,y)???, including a value for ???h(y)??? or ???h(x)???.
Set ???\Psi(x,y)=c??? to get the implicit solution.
How to solve an exact differential equation
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Step-by-step solution for finding the solution to the exact differential equation
Example
If the differential equation is exact, find its solution.
???x^3y^4+(x^4y^3+2y)\frac{dy}{dx}=0???
First, we’ll test to see whether or not the differential equation is exact. Matching the given equation to the standard form of an exact differential equation, we can say that
???M(x,y)=x^3y^4???
and
???N(x,y)=x^4y^3+2y???
We’ll test to see whether ???M_y=N_x???.
???\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}???
???4x^3y^3=4x^3y^3???
Both integrals will work, so we should look at M and N and then choose whichever function will be easier to integrate.
Since ???M_y=N_x???, we know that the given differential equation is an exact differential equation. Now we just need to find the solution ???\Psi(x,y)=c???.
???M(x,y)??? and ???N(x,y)??? are equally easy to integrate, we’ll just use ???M(x,y)??? and the integral
???\Psi=\int M(x,y)\ dx???
???\Psi=\int x^3y^4\ dx???
???\Psi=\frac14x^4y^4+h(y)???
We had to add ???h(y)??? instead of just ???C??? because we integrated a multivariable function with respect to ???x??? only, which doesn’t account for the integration of ???y???. Now we need to find ???h(y)???, which we’ll do by taking the partial derivative of both sides of the equation with respect to ???y???.
???\frac{d}{dy}\Psi=\frac{d}{dy}\left[\frac14x^4y^4+h(y)\right]???
???\Psi_y=x^4y^3+h\prime(y)???
Because we know that ???\Psi_y=N(x,y)???, we’ll make that substitution and then solve for ???h\prime(y)???.
???x^4y^3+2y=x^4y^3+h\prime(y)???
???2y=h\prime(y)???
To find ???h(y)???, we’ll integrate both sides with respect to ???y???.
???\int2y\ dy=\int h\prime(y)\ dy???
???y^2+k_1=h(y)+k_2???
???h(y)=y^2+k_1-k_2???
???h(y)=y^2+k???
Plugging this into the equation for ???\Psi??? gives
???\Psi=\frac14x^4y^4+h(y)???
???\Psi=\frac14x^4y^4+y^2+k???
Finally, setting ???\Psi=c??? to find the solution to the exact differential equation, we get
???c=\frac14x^4y^4+y^2+k???
???c-k=\frac14x^4y^4+y^2???
???c=\frac14x^4y^4+y^2???
We can say that the equation is exact and that its general (implicit) solution is
???c=\frac14x^4y^4+y^2???
Sometimes we’ll be given an initial condition and asked to find an explicit solution, instead of a general (implicit) solution. If that’s the case, we just plug the initial condition into the solution to find a value for ???c???. For example, if the previous example had given the initial condition ???y(0)=2???, our solution would have been
???c=\frac14x^4y^4+y^2???
???c=\frac14(0)^4(2)^4+(2)^2???
???c=4???
???4=\frac14x^4y^4+y^2???
