Differential Equations: Exact Equations

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Exact Differential Equations ....

Properties:

* .... know what they are right off hand.

* .... have method that we know to solve these equations.

First order equations can be written of the form: M(x,y)dx + N(x,y)dy = 0 (Krantz) Scientist like to find simpler ways to solve problems, and why shouldn't one method of problem solving help with another. Remember learning about circles in algebra, and that they had the form a^2 + b^2 = r^2? With r referring to radius, a family of curves (circles in this case) is designated by a specific radius. Well, it turns out that, using differential equations those families of curves can be used in such a way to solve problems.

So, what does all that mean on the above page. Well, given a function f(x,y) we take the partial derivative with respect to x and designate that as M, while also taking the partial derivative with respect to y and designating that as N. A partial derivative is taken when we hold the other variable constant during differentiation. Thus, x is held constant when (partial) differentiating with respect to y, and y is held constant when (partial) differentiating with respect to x. Then, substituting M and N into the original equation we get the result on the second to last line above. The condition of the equation then implies that each partial derivative must be equal to zero. Continuing below, this condition can only be true if the function f(x,y) is a constant. But, remember our circles? That means that the resulting differential equation is like a family of curves!!!!

Looking above we then see that a second partial derivative of the function is taken with respect to the OTHER variable. This action then leads to the Exactness condition as seen.

Now, let us see an example of how Exactness is used!

We see that the given equation is not separable, but that after manipulation (sorry for the messiness) it clearly has an M and an N (placing it in the form M(x,y) + N(x,y) = 0). Now, however, it can be checked for exactness taking partial derivatives on both sides. As can be seen in this form it is not exact. Looking below, however, after multiplying both sides by x^2 the story is quite different. Taking partial derivatives of both sides now results in a positive test for exactness. The exercise then is find f(x,y) by taking partial derivatives of f(x,y) with respect to both x and y.

First doing so with respect to x and then integrating yields the f(x,y) with an added function of y. Then, the whole function is differentiated with respect to y yielding N(x,y) with a derivative of the function. Thus, integrating the new function portion yields a new constant, C1. (bottom below)

Finally, remembering f(x,y) = C (family of curves) the new constant can be added to the original f(x,y) = -x^3 cos y + C1. As can be seen our final solution is derived from solving what's left for y.

You can go to DE Ex2 here...D.E. Ex2.

You can go to DE Ex1 here ...D.E. Ex1.