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 in the form M(x, y) dx + N(x, y) dy = 0, where there exists a function F(x, y) such that dF = M dx + N dy, meaning that the mixed partial derivatives of F are equal. When an equation is exact, it implies that there is a potential function F whose total differential gives back the original differential equation, and solving the equation reduces to finding F(x, y) = C, where C is an arbitrary constant.

Test for Exactness

The test for exactness involves checking whether the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x, that is, if M_y = N_x. This condition is necessary and sufficient for the differential equation to be exact if M and N have continuous second partial derivatives. This test helps in determining the appropriate solution method for the differential equation.

Integrating Factor

An integrating factor is a function, typically dependent on x, y, or both, that when multiplied by a non-exact differential equation, renders it exact. Integrating factors are particularly useful when the exactness test fails. Methods for determining integrating factors include inspection and the application of theorems that suggest forms of integrating factors depending on the structure of the differential equation.

Initial Conditions and Particular Solutions

Initial conditions are additional constraints provided with a differential equation that allow for the determination of a unique particular solution from a family of general solutions. After obtaining the general solution, the initial condition is applied to resolve any constants of integration, thus yielding the specific solution that satisfies the given condition.

Transcript

00:01 We have the given equation of the form.

00:06 You have m.

00:08 X, y, the x, plus n.

00:16 You have x, y, to be equal to zero.

00:24 So, therefore, we have x minus y, the x, minus the y to be equal to zero so let's simplify this this is going to give us x minus y the x you have minus x minus y the y to be equal to zero so we want to write it in this form so i have x minus y the x plus x minus y the y to be x plus x minus y so this will give us y so this will give us y y minus x d y to be equal to zero so what happens you realize that this becomes our aim and this becomes n so now we can find the partial derivative so you have the partial derivative of m with respect to y which should give you negative 1 then the partial derivative of n with respect to x which is also equal to negative 1.

02:11 So then they are exact.

02:15 So then this implies that your partial derivative of m with respect to y, it's equal to your partial derivative of n with respect to x.

02:30 So equation, this equation then is exact.

02:38 And because it is exact we cannot calculate the general solution based on this formula so let's calculate our general solution where your general solution then will be calculated based on you to be put to the integral you you have the x plus your constants which is k y so therefore u is going to be equal to the integral of m is x minus y the x plus your constant k y.

03:39 So you have you to be equal to the integral of this will give us x squared divided by two minus x y plus your constant k y.

03:59 So to calculate for k y so to calculate for this we need to differentiate this formula with respect to why so we differentiate partial derivative of you with respect to why and this will give us n so this there is going to be so you have if you have if you different with respect to n, this is going to give you, so you have negative x.

04:58 So this will give us negative x plus the k divided by the y.

05:09 And our n is y minus x so what do you realize you realize that you have this so you have this and that so this implies that your decay the y is equal to y which implies that decay it's equal to y, d, y...

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