Solution of Exact Equations

Contents

• First order ordinary differential equation

• Differential of a function of two variables

• Short Notes on Partial Derivatives

• Exact Equations

• Criterion for Exactness

• Examples

• Method of Solution

• Worked Example

• Practice Problems

• Solutions to practice problems

First Order Ordinary differential equations • A differential equation having a first derivative as the highest

derivative is a first order differential equation.

• If the derivative is a simple derivative, as opposed to a partial derivative, then the equation is referred to as ordinary.

Differential of a function of two variables If given a function 𝑧 = 𝑓(𝑥, 𝑦)

Then its differential is defined as the following:

𝑑𝑧 =𝜕𝑓

𝜕𝑥𝑑𝑥 +

𝜕𝑓

𝜕𝑦𝑑𝑦

The symbol 𝜕 represents the partial derivative of the function.

This tells us that if we know the differential of a function, we can get back the original function under certain conditions.

In a special case when 𝑓 𝑥, 𝑦 = 𝑐 then we have: 𝜕𝑓

𝜕𝑥𝑑𝑥 +

𝜕𝑓

𝜕𝑦𝑑𝑦 = 0

Short note on Partial derivatives For a function of two variables, a partial derivative with respect to a particular variable means differentiating the function with that variable while assuming the other variable to be fixed.

Ex: if 𝑓 = 𝑥 + 𝑦 sin 𝑥 + 𝑦3𝑙𝑛𝑥

then 𝜕𝑓

𝜕𝑥= 1 + 𝑦 cos 𝑥 + 𝑦3

1

𝑥

and 𝜕𝑓

𝜕𝑦= sin 𝑥 + 3𝑦2𝑙𝑛𝑥

Exact Equation

If given a differential equation of the form 𝑀 𝑥, 𝑦 𝑑𝑥 +𝑁 𝑥, 𝑦 𝑑𝑦 = 0

Where M(x,y) and N(x,y) are functions of x and y, it is possible to solve the equation by separation of variables.

However, another method can be used is by examining exactness.

The whole idea is that if we know M and N are differentials of f, then it is possible to reconstruct the original function f.

Criterion for Exactness

This method can only be used if the differential given is exact. Specifically,

𝜕𝑀

𝜕𝑦=𝜕𝑁

𝜕𝑥

If the above condition is satisfied, then we can use the method.

Remember, the function M(x,y) is the one that multiplies dx and the function N(x,y) is the one that multiplies dy.

Look at some of the examples on the next slide

Examples

2𝑥𝑦𝑑𝑥 + 𝑥2 − 1 𝑑𝑦 = 0

Here,

𝑀 𝑥, 𝑦 = 2𝑥𝑦 and 𝑁 𝑥, 𝑦 = 𝑥2 − 1

Then, we have 𝜕𝑀

𝜕𝑦= 2𝑥 and

𝜕𝑁

𝜕𝑥= 2𝑥

Therefore, this particular equation is exact.

Another Example

𝑒2𝑦 − 𝑦 cos 𝑥𝑦)𝑑𝑥 + (2𝑥 𝑒2𝑦 − 𝑥 cos 𝑥𝑦 + 2𝑦 𝑑𝑦 = 0

Here, 𝑀 𝑥, 𝑦 = 𝑒2𝑦 − 𝑦 cos 𝑥𝑦 ; 𝑁 𝑥, 𝑦 = 2𝑥𝑒2𝑦 −𝑥 cos 𝑥𝑦 + 2𝑦

Therefore, 𝜕𝑀

𝜕𝑦= 2𝑒2𝑦 − cos 𝑥𝑦 + 𝑥𝑦 sin 𝑥𝑦 =

𝜕𝑁

𝜕𝑥

As we see, this equation is also exact.

Method of Solution

The basic idea behind the obtaining the solution is very simple really. It’s also logical.

• If given a differential, we first test for exactness. If it is exact,

we know that since 𝜕𝑓

𝜕𝑥= 𝑀 𝑥, 𝑦 then 𝑓 = 𝑀 𝑥, 𝑦 𝑑𝑥 +

ℎ(𝑦)

• Since integration yields a constant of integration, the most general constant would be a function of y, since on differentiating it with x, we get 0.

Contd.

• Since we know that the reconstructed function must satisfy the condition

𝜕𝑓

𝜕𝑦= 𝑁(𝑥, 𝑦) then 𝑁 𝑥, 𝑦 =

𝜕

𝜕𝑦 𝑀 𝑥, 𝑦 𝑑𝑥 + ℎ 𝑦

• From this, we will be able to solve the above expression for h’(y). • Integrating that, we get h(y) and our solution is complete. • The complete and most general solution is then

𝑀 𝑥, 𝑦 𝑑𝑥 + ℎ 𝑦 = 𝑐

Worked Example

Given 𝑑𝑦

𝑑𝑥=

𝑥𝑦2−sin 𝑥 cos 𝑥

𝑦(1−𝑥2)

As we can see, this would be extremely difficult to solve using separation of variables. However, let’s try and use exactness to solve the equation. Rearranging it, we get

𝑥𝑦2 − sin 𝑥 cos 𝑥 𝑑𝑥 − 𝑦 1 − 𝑥2 𝑑𝑦 = 0

Testing for exactness, 𝜕𝑀

𝜕𝑦= 2𝑥𝑦 =

𝜕𝑁

𝜕𝑥

Thus, we know the equation is exact, so we proceed with our method.

Contd.

Now, we know that integrating M(x,y) with respect to x will give us the original function back.

Thus, (𝑥𝑦2 − sin 𝑥 cos 𝑥)𝑑𝑥 =𝑥2𝑦2

2+

cos 2𝑥

4+ ℎ(𝑦)

Furthermore, 𝜕

𝜕𝑦

𝑥2𝑦2

2+cos 2𝑥

4+ ℎ(𝑦) = 𝑥2𝑦 + ℎ′(𝑦)

And this should be equal to N(x,y).

Contd.

Then we have, 𝑥2𝑦 + ℎ′ 𝑦 = −𝑦(1 − 𝑥2)

Simplifying this, we get ℎ′ 𝑦 = −𝑦 Notice that if the equation was not exact to begin with, then h’(y) will not be a function of y alone in the step above. Solving for h(y), we have

ℎ 𝑦 = −𝑦2

2

Hence, the complete solution is 𝑥2𝑦2

2+cos 2𝑥

4−𝑦2

2=c

Where ‘c’ is an arbitrary constant.

Practice Problems

Determine Whether these equations are exact or not. If they are, solve them.

1. 𝑥𝑑𝑦

𝑑𝑥= 2𝑥𝑒𝑥 − 𝑦 + 6𝑥2

2. (𝑥2−𝑦2)𝑑𝑥 + (𝑥2 − 2𝑥𝑦)𝑑𝑦 = 0

3. 5𝑦 − 2𝑥 𝑦′ − 2𝑦 = 0

Answers to Practice Problems

1. 𝑥𝑦 − 2𝑥𝑒𝑥 + 2𝑒𝑥 − 2𝑥3 = 𝑐

2. Not Exact. (𝜕𝑀

𝜕𝑦= −2𝑦 ≠ 2𝑥 − 2𝑦 =

𝜕𝑁

𝜕𝑥)

3. Not Exact (𝜕𝑀

𝜕𝑦= 5 ≠ 0 =

𝜕𝑁

𝜕𝑥)

References

• A first Course in Differential Equations 9th Ed., Dennis Zill.

• Fundamentals of Differential Equations 3rd Ed, Nagle & Saff

• Differential and Integral Calculus Vol 2, N. Piskunov

Good Luck!