First-Order Differential Equations Part 2: Exact & Homogeneous Types
Differential of a Function of Two Variables Recall that if z = f(x, y) is a function of two variables with continuous first derivatives in a region R of the xy-plane, then its differential is In the special case when f(x, y) = c, where c is a constant, then
Differential of a Function of Two Variables In other words, given a one-parameter family of functions f(x, y) = c, we can generate a first-order differential equation by computing the differential of both sides of the equality.
Differential of a Function of Two Variables For example:
Differential of a Function of Two Variables Of course, not every 1st-order differential equation written in differential form M(x, y)dx + N(x, y)dy = 0 corresponds to a differential of f(x, y) = c
Exact Equation A differential expression M(x, y)dx + N(x, y)dy is an exact differential in a region R of the xy-plane if it corresponds to the differential of some function f(x, y) defined in R. A first-order differential equation of the form M(x, y)dx + N(x, y)dy = 0 is said to be an exact equation if the expression on the left-hand side is an exact differential.
Illustration x2y3dx + x3y2dy = 0 is an exact equation because the left-hand side is an exact differential of (1/3)x3y3: M(x, y) = x2y3 N(x, y) = x3y2
Illustration Notice also that for M(x, y) = x2y3 N(x, y) = x3y2 we have
Criterion for an Exact Differential Let M(x, y) and N(x, y) be continuous and have continuous first partial derivatives in a rectangular region R defined by a < x < b, c < y < d. Then a necessary and sufficient condition that M(x, y)dx + N(x, y)dy be an exact differential is
Solving Exact Equations Step 1. Determine if M(x, y)dx + N(x, y)dy = 0 is exact by checking if
Solving Exact Equations Step 2. If the equation is exact, find f(x, y) from either M(x, y) or N(x, y). If we use M(x, y), then assuming y is constant, g(y) is the “constant” of integration.
Solving Exact Equations Step 3. Differentiate the f(x, y) from step 2 with respect to the other variable y: But the left side of the equation is N(x, y).
Solving Exact Equations Step 4. Solve for g’(y).
Solving Exact Equations Step 5. Integrate the equation g’(y) with respect to y and substitute the resulting g(y) into The implicit solution of the equation is f(x, y) = c.
Step 1: Check if ∂M/∂y = ∂N/∂x Be careful about the signs Step 1: Check if ∂M/∂y = ∂N/∂x
Step 3: Differentiate F with respect to the other variable. Step 2: Find F. Step 3: Differentiate F with respect to the other variable.
The g’(y) of the last equation on the previous page is equal to the encircled quantity
Step 4: Solve for either g’(y) or g’(x), whichever is applicable. In other words, Step 4: Solve for either g’(y) or g’(x), whichever is applicable.
Step 5: Integrate either g’(y) or g’(x), whichever is applicable, then substitute into F. Hence,
Solution by Substitutions We usually solve a differential equation by recognizing it as a certain kind of equation (say, separable, linear, exact) and then carrying out a procedure, consisting of equation-specific mathematical steps, that yields a solution of the equation.
Solution by Substitutions But it is not uncommon to be stumped by a differential equation because it does not fall into one of the classes of equations previously discussed. Now, there are three different kinds of first-order differential equations that are solvable by means of a substitution. Homogeneous Equations Bernoulli’s Equation Reduction to Separation of Variables
Homogeneous Equations If a function f possess the property f ( tx, ty ) = t f ( x, y ) for some real number , then f is said to be a homogeneous function of degree . For example, f(x, y) = x3 + y3 is a homogeneous function of degree 3 since f(tx, ty) = (tx)3 + (ty)3 = t3(x3 + y3)
Homogeneous Equations A first-order DE in differential form M(x, y)dx + N(x, y)dy = 0 is said to be homogeneous if both coefficient functions M and N are homogeneous equations of the same degree. In other words, M(tx, ty) = tM(x, y) N(tx, ty) = tN(x, y)
Homogeneous Equations In addition, if M and N are homogeneous functions of degree , we can also write M(x, y) = xM(1, u) N(x, y) = xN(1, u) M(x, y) = yM(v, 1) N(x, y) = yN(v, 1) where y = ux where x = vy
Homogeneous Equations Either of the substitutions y = ux or x = vy where u and v are new dependent variables, will reduce a homogeneous equation to a separable first-order differential equation.
Homogeneous Equations In practice, for M(x,y)dx + N(x,y)dy = 0, we use y = ux if N(x,y) is simpler than M(x,y), or we use x = vy if M(x,y) is simpler than N(x,y).
Solution: Since the equation is homogeneous:
Solution:
Solution:
Solution:
Solution: Since the equation is homogeneous:
Note on the substitution Although either y = ux or x = vy can be used for every homogeneous differential equation, in practice we try x = vy whenever the function M(x, y) is simpler than N(x,y). Also it could happen that after using one substitution, we may encounter integrals that are difficult or impossible to evaluate in closed form; switching substitutions may result in an easier problem.