Review on partial derivatives and Chain Rules for functions of 2 variables.
Consider a function f (x, y). We use these notations for the partial derivatives:
and correspondingly the higher derivatives:
and the cross derivatives
where we have the identity
Now, consider x = x(t) and y = y(t), and we form a composite function as f (x(t), y(t)).
We see that f now depends only on t.
Example 1. Let f (x, y) = x2y2 + ex, and x(t) = t2, y(t) = et , and consider the composite function f (x(t), y(t)). Compute
Answer. We ﬁrst compute the derivatives
By the Chain Rule, we compute
Special case: If y = y(x), then the composite function f (x, y(x)) will follow this form of Chain Rule
Example 2. Let y(x) be the unknown. Consider the equation
6x + exy2 + 2exyy′ = 0
We see that the equation is NOT linear. It is NOT separable either. None of the methods we know can solve it.
However, deﬁne the function
ψ(x, y) = 3x2 + exy2
We notice that
ψx = 6x + exy2, ψy = 2ex y and the equation can be written as
ψx (x, y) + ψy (x, y)y′ = 0.
Since y = y(x), we apply the Chain Rule to the composite function ψ(x, y(x)) and get
which is the left-hand side of the equation. By the diﬀerential equation, we now have
where C is an arbitrary constant, to be determined by initial condition. We have found the solution in an implicit form :
3x2 + exy2 = C.
In this example, we are even able to write out the solution in an explicit form by algebraic manipulation
Here, the choice of + or − sign should be determined by initial condition.
Deﬁnition of an exact equation. En equation in the form
M (x, y) + N (x, y)y′ = 0
is called exact if there exists a function ψ(x, y) such that ψx = M and ψy = N .
How to check if an equation is exact? By the identity ψxy = ψyx, we must have My (x, y) = Nx(x, y).
How to solve it? Need to ﬁnd the function ψ, then we get implicit solution
ψ(x, y) = C.
How to ﬁnd ψ? By using the facts that
ψx = M (x, y), ψy = N (x, y) and integrate. We will see this through an example.
Example 3. Check if the following equation is exact (2x + 3y) + (x − 2y)y′ = 0.
Answer. Here we have
M (x, y) = 2x + 3y, N (x, y) = x − 2y.
My = 3, Nx = 1, My ≠ Nx,
so the equation is not exact.
Remark 1. Consider a separable equation:
Multiply both sides by g(y) and re-arranging terms, we get
f (x) − g(y)y′ = 0.
Now we check if this equation is exact. Clearly, we have fy = 0 and gx = 0, so it is exact. We may conclude that all separable equations can be rewritten into an exact equation.
Remark 2. However, the exactness of an equation is up to manipulation. Consider the separable equation in Remark 1, and rewrite the equation now into
So now we have
Since Nx = 0 and My = 0 in general, the equation is not exact.
So, be careful. When you say an equation is exact, you must specify in which form you present your equation.
Example 4. Consider the equation
(2x + y) + (x + 2y)y′ = 0
(1) Is it exact? (2) If yes, ﬁnd the solution with the initial condition y(1) = 1.
Answer.(1). We have M = 2x + y and N = x + 2y, so My = 1 and Nx = 1, so the equation is exact.
(2). To solve it, we need to ﬁnd the function ψ.
We have ψx = 2x + y, ψy = x + 2y. (A)
Integrating the ﬁrst equation in x:
(Review: To perform a partial integration in x, we treat y as a constant. Therefore, the integration constant could depend on y since it is a constant. That’s why we add h(y), a function of y, to the anti-derivative.)
To determine h(y), we use the second equation in (A).
ψy = x + h′ (y) = x + 2y, h′ (y) = 2y, h(y) = y2.
ψ = x2 + xy + y2
and the implicit solution is
x2 + xy + y2 = C.
Finally, we determine the constant C by initial condition. Plug in x = 1, y = 1, we get C = 3, so the implicit solution is
x2 + xy + y2 = 3.
Example 5. Given equation
(xy2 + bx2 y) + (x + y)x2 y′ = 0. (1)
Find the values of b such that the equation is exact. (2) Solve it with that value of b.
Answer.(1). We have
M (x, y) = xy2 + bx2y, N (x, y) = x3 + x2y
My = 2xy + bx2, Nx = 3x2 + 2xy
We see that we must have b = 3 to ensure My = Nx which would make the equation exact. (2). We now set b = 3. To solve the equation, we need to ﬁnd the function ψ. We have
ψx = xy2 + 3x2y, ψy = x3 + x2y.
Integrating the ﬁrst equation in x:
To ﬁnd h(y), we use ψy :
ψy = x2y + x3 + h′(y) = Nx = 3x2 + 2xy
We must have h′ (y) = 0, so we can use h(y) = 0.
The implicit solution is
where C is arbitrary, to be determined by initial condition.