solve the subsequent differential equation2xy - 9

Question

Solve the subsequent differential equation.

2xy - 9 x² + (2y + x² + 1) dy/dt = 0

Solution

Let's start off via supposing that wherever out there in the world is a function Ψ(x,y) which we can get. For this illustration the function that we require

Ψ(x,y) = y² + (x² + 1) y - 3x³

Do not worry at this point about where this function came from and how we found it.  Finding the function, ?(x,y), that is needed for any particular differential equation is where the vast majority of the work for these problems lies.  As stated earlier, however, the point of this example is to demonstrate you why the solution process works rather than demonstrating you the actual solution process.

We will notice how to find this function in the subsequent example, thus at this point do not worry regarding how to find it, eailsy accept that this can be got and that we've done as for this exact differential equation.

Here, take some partial derivatives of the function.

Ψ_x = 2xy - 9x²

Ψ_x = 2y + x² + 1

Here, compare these partial derivatives to the differential equation and you'll observe that along with these we can here write the differential equation like:

Ψ_x + Ψ_x dy/dt = 0    ......................(1)

Currently, recall from your multi-variable calculus class that (1) is nothing more than the subsequent derivative which you'll require the multi-variable chain rule for this.

d/dx(Ψ(x,y(x)))

Thus, the differential equation can now be written like:

d/dx(Ψ(x,y(x))) = 0

There, if the ordinary but not partial derivative of something is zero, as something should have been a constant to start along with. Conversely, we've got to contain Ψ(x,y) = c.  Or,

y² + (x² + 1) y - 3x³ = c

This afterward is an implicit solution for our differential equation! If we had a first condition we could solve for c. We could also get an explicit solution if we needed.

Okay, therefore what did we learn from the last illustration? Let's consider things a little more commonly. Assume that we have the subsequent differential equation.

M(x,y) + N (x,y) dy/dx = 0   .......................(2)

Remember that it's significant that it be in this form! There should be an "= 0" on individual side and the sign separating the two terms should be a "+".  There, if there is a function somewhere out there in the world, Ψ (x,y), hence,

Ψ_x = M(x,y) and

Ψ_x = N (x,y)

After that we call the differential equation accurate. In such cases we can write the differential equation like:

Ψ_x + Ψ_x dy/dt = 0    ......................(3)

After that using the chain rule from Calculus III we can further decrease the differential equation to the subsequent derivative,

d/dx(Ψ(x,y(x))) = 0

The implicit solution to an accurate differential equation is after that:

Ψ(x,y) = c   ......................(4)

Well, it's the solution given we can get Ψ (x,y) anyway. Thus, once we have the function we can all the time just jump straight to (4) to find an implicit solution to our differential equation.

Finding the function Ψ (x,y) is obviously the central task in finding if a differential equation is accurate and in finding its solution. When we will notice, finding Ψ (x,y) can be a somewhat lengthy process wherein there is the chance of mistakes. Thus, it would be nice whether there was some easy test that we could use before even starting to notice if a differential equation is exact or not. It will be especially helpful if it turns out that the differential equation is not exact, as in this case Ψ (x,y) will not exist. This would be a waste of time to try and get a nonexistent function!

Therefore, let's see if we can get a test for exact differential equations. Let's begin with (2) and suppose that the differential equation is in fact exact. As it's accurate we know that somewhere out there is a function Ψ (x,y) which satisfies,

Ψ_x = M

Ψ_y = N

Currently, provided Ψ(x,y) is continuous and its first order derivatives are as well continuous we identify that

Ψ_x,y = Ψ_y,x

Though, we also have the subsequent.

Ψ_x,y = (Ψ_x)_y = (M)_y = M_y

Ψ_y,x = (Ψ_y­) _x  = (N)_x = N_x

Thus, if a differential equation is exact and Ψ (x,y) meets all of its continuity conditions we should have as:

M_y  = N_x  .......................(5)

Similarly if (5) is not true there is no way for the differential equation to be exact.

Thus, we will use equation (5) as a test for exact differential equations. Whether (5) is true we will suppose that the differential equation is exact and that Ψ(x,y) meets all of its continuity conditions and proceed along with determining it. Remember that for all the illustrations here the continuity conditions will be met and thus this won't be an issue.