1. Chapter 2.6: Exact equations and integrating factor. In some cases the equation M (x, y) + N (x, y)y ′ = 0 ⇔ M (x, y)dx + N (x, y)dy = 0 is not exact, but it can be modified to become exact. More precisely, the condition My = Nx is not satisfied. We seek for an integrating factor µ(x, y) such that the equation µM dx + µN dy = 0 is exact, that is (M µ)y = (N µ)x . Once such a µ is found, we can then solve the exact equation. How do we solve for µ? We expand (M µ)y = (N µ)x using the product rule µy M + µMy = µx N + µNx . This is a PDE, potentially more complicated than the original equation. To simplify it, we try µ = µ(x), that is independent µ is independent of y; in particular µx = 0. Then the equation becomes Nx − My . µMy = µx N + µNx ⇔ µx N = µNx − µMy ⇔ µx = µ N N −M This can be solved only if x N y does not depend on y! If that is the case, we need to solve a separable/linear equation. Example. Solve y + (2xy − e−2y )y ′ = 0 ⇔ ydx + (2xy − e−2y )dy = 0. We identify M = 2xy − e−2y , N = y. We compute My = 1 6= 2y = Nx . We seek such that (M µ)y = (N µ)x . We try µ = µ(x). We obtain µx = µ Nx − My 2y − 1 =µ ; N 2xy − e−2y 2y−1 this is not satisfactory since the expression 2xy−e −2y depends on y. We try µ = µ(y). Since the computations were not performed above, we do them from scratch: (M µ)y = (N µ)x ⇔ µy M +µMy = µNx +µx N ⇔ µy M +µMy = µNx ⇔ yµy +µ = 2yµ. We obtain: µy = which is fine since 2y−1 y 2y − 1 µ, y depends only on y and not on x. We separate it: 2y − 1 dµ = dy, µ y 1 2 keeping in mind that here the derivatives are taken with respect to y. We integrate Z Z Z 2y − 1 1 dµ = dy = 2 − dy ⇔ ln |µ| = 2y − ln |y|. µ y y Thus e2y . µ = e2y−ln y = y Note that we did not care about constants and drop the absolute value: we can do this since we do not solve for general µ; a single one does the job. Now we go back to the original equation and multiply it with µ: 1 e2y dx + (2xe2y − )dy = 0. y It is clear that 1 (e2y )y = 2e2y = (2xe2y − )x . y We seek for ψ such that 1 ψx = e2y , ψy = (2xe2y − ). y Let’s start with the first equation: Z ψ = e2y dx = xe2y + C(y). We plug this into the other equation: Z 1 1 1 2y ′ 2y ′ 2xe + C (y) = 2xe − ⇔ C (y) = − ⇔ C(y) = − = − ln |y| + C. y y y Thus ψ(x, y) = xe2y − ln |y| and the general solution is given by xe2y − ln |y| = C.
