Exact equations are first order DEs of the form M(x, y) + N(x, y) y' = 0 for which we can find a function f(x, φ(x)) so that

Transcription

1 Section 2.6 Exact Equations (ONLY) Key Terms: Exact equations are first order DEs of the form M(x, y) + N(x, y) y' = 0 for which we can find a function f(x, φ(x)) so that The construction of f(x, φ(x)) will involve partial derivatives and the solution of some related DEs

2 Review of a Calculus Idea: Let f(x,y) be a function of two variables, where y = y(x), that is, y is some function of x. The derivative of f(x,y) with respect to x is computed as follows: Expressed in differential form we have this is called the total differential of function f(x,y). The only way the total differential can equal zero is f(x,y) is a constant function.

3 Examples: cos( xy) cos( xy) For f(x,y) = cos(xy) we have df dx dy x y xy xy df sin( xy) dx sin( xy) dy sin( xy) ydx sin( xy) x dy x y x y x y ye ye 2 2 df dx dy x y

4 Some ODEs can be solved by trying to view the DE as a total differential of a function and then constructing that function. Those for which we can find the function are called exact DEs. Example: and in differential form as If we let f(x,y) = xy we see that df = ydx + xdy so df = d(xy) = 0. The only way the differential of a function equals zero is if it is a constant. Thus the implicit expression xy = C is a family of solutions to the original DE. (This equation is also separable.) Unfortunately it is not always that easy to find the function f(x,y) whose differential is the differential rearrangement of the DE. DEFINITION: A first order ODE y' = G(x,y) is called an EXACT ODE if it can be rearranged into the form M(x,y) dx + N(x,y) dy = 0 and there exists a function f(x,y) such that the exact differential of f(x,y) is M(x,y) dx + N(x,y) dy, that is df(x,y) = M(x,y) dx + N(x,y) dy If the ODE is exact, then f(x,y) = C is an implicit solution of the ODE.

5 What we need is an easy way to recognize EXACT ODEs. By definition the total differential of f is f f df(x, y) dx dy x y Luckily this is given to us in the form of a Calculus check! If df(x,y) = M(x,y) dx + N(x,y) dy then it must be that Recall that Recall that if f(x,y) has continuous second order partial derivatives, then the mixed partial derivatives must be equal. That is, Working with this relationship we do the following. Since we have TEST for EXACTNESS: ODE M(x,y) dx + N(x,y) dy = 0 is exact if and only if The mixed partials are equal.

6 Examples: Determine if the following equations are exact. Rearrange into the standard form M(x,y)dx + N(x,y)dy = 0 Use the test for exactness M y =0 N x = 1 So the DE is not Exact. Rearrange into the form M(x,y)dx + N(x,y)dy = 0 Use the test for exactness

7 Once we have identified an exact equation, how do we solve it? ANSWER: We construct f(x,y) so that f x = M and f y = N. To do this we integrate either f x = M or f y = N partially. The procedure of partial integration uses an arbitrary function in the unknown that was held fixed in the integration process, rather than just a constant of integration. For example, starting with f x = M we integrate partially with respect to x and write then use the fact that f y = N to determine h(y). The alternate procedure: Using f N(x,y) integrate partially with respect to y; then y Here we integrate partially with respect to x so the constant must be a function of y. f fydy N(x, y)dy h(x) Here we integrate partially with respect to y so the constant must be a function of x. Then use fx M(x, y) to find h(x). Since df = M(x,y)dx + N(x,y)dy = 0, it must be that the expression f(x,y) = C is the general solution of the DE M(x,y)dx + N(x,y)dy = 0.

8 Example: Solve exact DE Start with f x = M; integrate partially with respect to x. Next use compute the partial with respect to y of the preceding expression and set it equal to N f y = N = x 2 f y 2 3 x y x h(y) y x 2 x dh x dy Simplify. So h'(y) = 0 h(y) = C a constant f = x 2 y+x 3 + C Since df(x,y) = (2xy+3x 2 )dx + x 2 dy = 0 where f = x 2 y+x 3 + C we have solution of the DE is f(x, y) = x 2 y + x 3 + C = K, a constant. Now we can write f(x, y) = x 2 y + x 3 = K C = C since a difference of two arbitrary constants is another constant.

9 More Examples: Integrate partially with respect to x. So DE is exact. Diifferentiate f with respect to y and use that f y = N. So is the solution. So DE is exact. Integrate partially with respect to x. Use that f y = N. So is the solution.

10 Summary: Given a DE in the form M(x,y)dx + N(x,y)dy = 0. If M N My is equal to Nx then the DE is called EXACT. y x In that case we can find a function f(x,y) so that the exact differential of f(x,y) is equal to M(x,y)dx + N(x,y)dy; that is, df(x,y) = M(x,y)dx + N(x,y)dy. Since the exact differential of f(x, y) is given by the expression f(x, y) f(x, y) we have that M(x, y) fxand N(x, y) fy x y From this we can construct a formula for f(x,y) using (partial) integration. Using fx M(x, y) integrate partially with respect to x; then f fxdx M(x, y)dx h(y) An alternate procedure starts with f N(x, y) y See the next page. Then use fy N(x, y) to find h(y). Since df = M(x,y)dx + N(x,y)dy = 0, it must be that the expression f(x,y) = C is the general solution of the DE M(x,y)dx + N(x,y)dy = 0.

11 The alternate procedure: Using fy N(x, y) integrate partially with respect to y; then f fydy N(x, y)dy h(x) Here we integrate partially with respect to y so the constant must be a function of x. Then use fx M(x, y) to find h(x). Since df = M(x,y)dx + N(x,y)dy = 0, it must be that the expression f(x,y) = C is the general solution of the DE M(x,y)dx + N(x,y)dy = 0.