Test for exactness. If exact, solve, If not, use an...

Key Concepts

Exact Differential Equations

An exact differential equation is one that can be written as the total differential of a function, meaning there exists a potential function F(x, y) such that its total derivative dF equals M(x, y) dx + N(x, y) dy. In this context, the equation can be integrated directly once it is confirmed to be exact, reducing the problem to finding F(x, y) whose partial derivatives match M and N.

Testing for Exactness

To determine if an equation is exact, the criterion ?M/?y = ?N/?x must be satisfied, where M and N are the coefficients of dx and dy, respectively. This test confirms whether a potential function exists for the differential equation, guiding the solution approach accordingly.

Integrating Factors

When a differential equation is found to be nonexact, an integrating factor—a function that depends on x, y, or both—can often be identified to multiply the equation and convert it into an exact one. This concept is crucial because it allows one to transform a difficult-to-solve equation into a form that is more manageable and amenable to integration.

Finding the Potential Function

Once exactness is established (either inherently or after applying an integrating factor), the next step is to recover the potential function F(x, y). This is done by integrating M with respect to x (or N with respect to y) and then ensuring consistency with the derivative with respect to y (or x) to account for any functions of the other variable. This method yields the general solution to the differential equation.

Initial Value Problems

When an initial condition is specified, it provides a particular value for the potential function at a given point, thereby determining the constant of integration. This turns the general solution into a particular solution that satisfies the additional constraint imposed by the initial condition.

Transcript

00:01 We have the standard form of a differential equation in the form.

00:08 So therefore we have e exponent x plus y minus y d s plus x plus x to be x to be equal to zero.

00:34 So then we notice that this x are n, then you have this to be a.

00:44 The only way to do is a first step is exact.

00:49 So to do that, you find a partial derivative of 1 with respect to y.

00:56 And then in d .e is the power x plus y, my next one.

01:03 Then you find a partial derivative of n with respect to x and this will give us p to the power x plus y so he noticed that the problem with respect to y is not equal to the partial derivative of n with respect to s so hence we are saying that the equation it's not exact and if it is not exact then we can calculate for the integrating factor so integrating factor integrating factor we are saying f should be equal to this so what is the formula we have the integral of r p x and a r here it's equal to one over k you have partial derivative of p with respect to y minus partial derivative of k o with respect to x so what are the values of p and q they add our m and n those are m and n so this is more like you have one divided by n partial derivative of p of m with respect to y minus partial derivative of n with respect to x and that's what xp to the power x plus one plus one then you have partial the negative of p with respect to yom with respect to y or was p to the power x plus y plus y 1 we have minus e to the power x plus y minus we have our x e to the power x plus y so i here would be equal to 1 divided by x e to the power x plus 1 plus 1 then this will give us 1 plus 1 divided by x e to the power x plus y this is equal to negative 1 so then this implies that we know r so our f of x is going to be e the integral of r r is negative 1 you realize that r here is negative 1 the x which is equal to e to the power negative x.

05:07 So now you multiply your integrating factor by the difference.

05:15 So you have no more like you have used to time over time.

05:19 Time the differential i use what defense is so then this implies that you have e to the power negative x you have e to the power x plus y minus y the x plus e to the power negative x we have x e to the power x plus e to the power negative x you have x e to the power x plus y plus one the y to be equal to zero so this is we give us b to the power y we have minus y e to the power negative x the x plus this times that to give us x x e to the power y plus e to the five negative x the y to be equal to zero this then becomes our new m and this becomes a new so let's take the partial derivative of m with respect to y...

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