Consider the general Differential Form of a First Order Differential Equation
M(x, y) dx + N(x, y) dy = 0.
Such an equation is Exact provided there exists a differentiable function Ψ(x, y) such that
Ψx(x, y) = M(x, y) and Ψy(x, y) = N(x, y).
The General Solution of an Exact Equation is given implicitly by
Ψ(x, y) = c,
where c is an arbitrary constant.
A necessary and — for simply connected domains also — sufficient condition for exactness is My(x, y) = Nx(x, y). Non-Exact Equations can be converted into exact equations by multiplying the coefficients by a suitable Integrating Factor µ = µ(x, y). Of particular interest are integrating factors of a special form, such as functions of x or y only, or functions of x + y or xy.
A. Which of the following is a test for the existence of an integrating factor µ = µ(x) that depends on x only? Circle your answer!
(i) (My − Nx)/N is a function of x only
(ii) (My − Nx)/M is a function of x only
(iii) (My − Nx)/(M − N) is a function of x only
(iv) (My − Nx)/(xM − yN) is a function of x only.
B. Use your answer in part A to show that the non-exact differential equation (4xy + 6y^2 ) dx + (x^2 + 4xy) dy = 0,
has an integrating factor of the form µ = µ(x). Find that integrating factor and use it to determine the general solution of this equation.