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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:

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) = x^{2}y^{2} + ex, and x(t) = t^{2}, 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^{x}y^{2} + 2e^{x}yy′ = 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) = 3x^{2} + e^{x}y^{2}

We notice that

ψ_{x} = 6x + e^{x}y^{2}, 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 :

3x^{2} + e^{x}y^{2 }= 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) = y^{2}.

Therefore

ψ = x^{2} + xy + y^{2}

and the implicit solution is

x^{2} + xy + y^{2} = 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

x^{2} + xy + y^{2} = 3.

**Example 5. Given equation**

(xy^{2} + bx^{2} y) + (x + y)x^{2} 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) = xy^{2} + bx^{2}y, N (x, y) = x^{3} + x^{2}y

so

M_{y} = 2xy + bx^{2}, Nx = 3x^{2 }+ 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 }= xy^{2} + 3x^{2}y, ψy = x^{3 }+ x^{2}y.

Integrating the ﬁrst equation in x:

To ﬁnd h(y), we use ψ_{y} :

ψ_{y} = x^{2}y + x^{3} + h′(y) = N_{x }= 3x^{2 }+ 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.

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

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