First Order Linear Differential Equations

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# First Order Linear Differential Equations - PowerPoint PPT Presentation

First Order Linear Differential Equations. Any equation containing a derivative is called a differential equation. A function which satisfies the equation is called a solution to the differential equation. The order of the differential equation is the order of the highest derivative involved.

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First Order Linear Differential Equations

Any equation containing a derivative is called a differential equation.

A function which satisfies the equation is called a solution to the differential equation.

The order of the differential equation is the order of the highest derivative involved.

The degree of the differential equation is the degree of the power of the highest derivative involved.

We can then integrate both sides.

This will obtain the general solution.

1

A first order linear differential equation is an equation of the form

To find a method for solving this equation, lets consider the simpler equation

Which can be solved by separating the variables.

(Remember: a solution is of the form y = f(x) )

or

Using the product rule to differentiate the LHS we get:

If we multiply both sides by

Returning to equation 1,

Now integrate both sides.

For this to work we need to be able to find

Note the solution is composed of two parts.

The particular solution satisfies the equation:

The complementary function satisfies the equation:

To solve this integration we need to use substitution.

If you look at page 113 you will see a note regarding the constant when finding the integrating factor. We need not find it as it cancels out when we multiply both sides of the equation.

The modulus vanishes as we will have either both positive on either side or both negative. Their effect is cancelled.

(note the shortcut I have taken here)

The modulus vanishes as we will have either both positive on either side or both negative. Their effect is cancelled.

(note the shortcut I have taken here)

So far we have looked at the general solutions only. They represent a whole family of curves. If one point can be found which lies on the desired curve, then a unique solution can be identified.

Such a point is often given and its coordinates are referred to as the initial conditions or values. The unique solution is called a particular solution.

Hence the particular solution is

Page 114 and 116 Exercise 1 and 2

TJ Exercise 1

Second-order linear differential equations

Note that the general solution to such an equation must include two arbitrary constants to be completely general.

Proof

and

Is a solution.

If we call these two values m1 and m2, then we have two solutions.

A and B are used to distinguish the two arbitrary constants.

The two arbitrary constants needed for second order differential equations ensure all solutions are covered.

The type of solution we get depends on the nature of the roots of this equation.

When roots are real and distinct

The auxiliary equation is

To find a particular solution we must be given enough information.

The auxiliary equation is

Using the initial conditions.

Solving gives

Thus the particular solution is

Page 119 Exercise 3 Questions 1(a), (b) 2(a), (b)

But we will now look at the solution when the roots are real and coincident.

Roots are real and coincident

When the roots of the auxiliary equation are both real and equal to m, then the solution would appear to be y = Aemx+ Bemx = (A+B)emx

A + B however is equivalent to a single constant and second order equations need two.

With a little further searching we find that y = Bxemx is a solution. Proof on p119

So a general solution is

The auxiliary equation is

Using the initial conditions.

Page 120 Exercise 4 Questions 1(a), (b) 2(a), (b)

But we will now look at the solution when the roots are complex conjugates.

Roots are complex conjugates

When the roots of the auxiliary equation are complex, they will be of the form

m1 = p + iq and m2 = p – iq. Hence the general equation will be

The auxiliary equation is

Substituting gives:

Substituting gives:

The particular solution is

Page 122 Exercise 5A Questions 1(a), (b) 2(a), (b)

TJ Exercise 2

Non homogeneoussecond order differential equations

Non homogeneous equations take the form

Suppose g(x) is a particular solution to this equation. Then

Now suppose that g(x) + k(x) is another solution. Then

Giving

From the work in previous exercises we know how to find k(x).

This function is referred to as the Complimentary Function. (CF)

The function g(x) is referred to as the Particular Integral. (PI)

General Solution = CF + PI

Finding the (CF): the auxiliary equation is

Substituting into the original equation

Finding the (CF): the auxiliary equation is

Substituting into the original equation

Finding the form of the PI

The individual terms of the CF make the LHS zero

The PI makes the LHS equal to Q(x). Since Q(x) ≠ 0, it stands to reason that the PI cannot have the same form as the CF.

When choosing the form of the PI, we usually select the same form as Q(x).

This reasoning leads us to select the PI according to the following steps.

• try the same form as Q(x)
• If this is the same form as a term of the CF, then try xQ(x)
• If this is the same form as a term of the CF, then try x2Q(x)

Finding the (CF): the auxiliary equation is

Substituting into the original equation

If the wrong selection of PI is made, you will generally be alerted to this by the occurrence of some contradiction in later work.

Page 126 – Exercise 7A as many as possible.

TJ Exercise 3. This assesses beyond poly and trig work.