Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

Physics for IIT JAM, UGC - NET, CSIR NET

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Physics : Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

The document Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
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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:

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

and correspondingly the higher derivatives:

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

and the cross derivatives

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

where we have the identity

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

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:

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

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)). ComputeLinear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

Answer. We first compute the derivatives

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

By the Chain Rule, we compute

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

Special case: If y = y(x), then the composite function f (x, y(x)) will follow this form of Chain Rule

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

 

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

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

which is the left-hand side of the equation. By the differential equation, we now have

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

where C is an arbitrary constant, to be determined by initial condition. We have found the solution in an implicit form :

3x2 + exy= C.

In this example, we are even able to write out the solution in an explicit form by algebraic manipulation

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

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.

Then

My = 3, Nx = 1, M≠ Nx,

so the equation is not exact.

Remark 1. Consider a separable equation:

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

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

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

So now we have

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

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:

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

(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.

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

My = 2xy + bx2, Nx = 3x+ 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

ψ= xy2 + 3x2y, ψy = x+ x2y.

Integrating the first equation in x:

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

To find h(y), we use ψy :

ψy = x2y + x3 + h′(y) = N= 3x+ 2xy

We must have h′ (y) = 0, so we can use h(y) = 0.

The implicit solution is

Linear Ordinary Differential Equations of First and Second Order (Part - 3), UGC - NET Physics Physics Notes | EduRev

where C is arbitrary, to be determined by initial condition.

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