Linear Ordinary Differential Equations of First and Second Order - 3 | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Exact Equations

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.

Chain Rule:

Example 1. Let f (x, y) = x²y² + ex, and x(t) = t², y(t) = e^t , 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 + e^xy² + 2e^xyy′ = 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 + e^xy²

We notice that

ψ_x = 6x + e^xy², ψ_y = 2e^x y and the equation can be written as

ψ_x (x, y) + ψ_y (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 + e^xy² = 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 ψ_x = M and ψ_y = N .

How to check if an equation is exact? By the identity ψ_xy = ψ_yx, we must have M_y (x, y) = N_x(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

ψ_x = 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.

Then

M_y = 3, N_x = 1, M_y ≠ N_x,

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 f_y = 0 and g_x = 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 N_x = 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 M_y = 1 and N_x = 1, so the equation is exact.

(2). To solve it, we need to find the function ψ.

We have ψx = 2x + y, ψ_y = 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).

ψ_y = x + h′ (y) = x + 2y, h′ (y) = 2y, h(y) = y².

Therefore

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

so

M_y = 2xy + bx², Nx = 3x² + 2xy

We see that we must have b = 3 to ensure M_y = N_x 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

ψ_x = xy² + 3x²y, ψy = x³ + x²y.

Integrating the first equation in x:

To find h(y), we use ψ_y :

ψ_y = x²y + x³ + h′(y) = N_x = 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.