9.5 Linear Equations

9.5 Linear Equations

Linear Equations

A first-order linear differential equation is one that can be put into the form

where P and Q are continuous functions on a given interval.

This type of equation occurs frequently in various sciences, as we will see.

Linear Equations

An example of a linear equation is xy′ + y = 2x because, for x ≠ 0, it can be written in the form

Notice that this differential equation is not separable

because it’s impossible to factor the expression for y′ as a function of x times a function of y.

Linear Equations

But we can still solve the equation by noticing, by the Product Rule, that

xy′ + y = (xy)′

and so we can rewrite the equation as (xy)′ = 2x

Linear Equations

If we now integrate both sides of this equation, we get

xy = x² + C or

If we had been given the differential equation in the form of Equation 2, we would have had to take the preliminary step of multiplying each side of the equation by x.

It turns out that every first-order linear differential equation can be solved in a similar fashion by multiplying both sides of Equation 1 by a suitable function I(x) called an

integrating factor.

Linear Equations

We try to find I so that the left side of Equation 1, when multiplied by I(x), becomes the derivative of the product I(x)y:

I(x)(y′ + P(x)y) = (I(x)y)′

If we can find such a function I, then Equation 1 becomes (I(x)y)′ = I(x) Q(x)

Linear Equations

Integrating both sides, we would have

so the solution would be

Linear Equations

To find such an I, we expand Equation 3 and cancel terms:

I(x)y′ + I(x)P(x)y = (I(x)y)′ = I′(x)y + I(x)y′

I(x) P(x) = I′(x)

This is a separable differential equation for I, which we solve as follows:

Linear Equations

where A = ±e^C. We are looking for a particular integrating factor, not the most general one, so we take A = 1 and use

Thus a formula for the general solution to Equation 1 is provided by Equation 4, where I is given by Equation 5.

Linear Equations

Instead of memorizing this formula, however, we just remember the form of the integrating factor.

Solve the differential equation

Solution:

The given equation is linear since it has the form of Equation 1 with P(x) = 3x² and Q(x) = 6x².

An integrating factor is

Example 1

Example 1 – Solution

Multiplying both sides of the differential equation by , we get

or

Integrating both sides, we have

cont’d

Example 2

Find the solution of the initial-value problem

x²y′ + xy = 1 x > 0 y(1) = 2

Solution:

We must first divide both sides by the coefficient of y′ to put the differential equation into standard form:

The integrating factor is

Example 2 – Solution

Multiplication of Equation 6 by x gives

Then

and so

cont’d

Example 2 – Solution

Therefore the solution to the initial-value problem is

cont’d

Application to Electric Circuits

Application to Electric Circuits

Here, we consider the simple electric circuit shown in Figure 4: An electro-motive force (usually a battery or generator) produces a voltage of E(t) volts (V) and a current of I(t) amperes (A) at time t.

Figure 4

Application to Electric Circuits

The circuit also contains a resistor with a resistance of R ohms (Ω) and an inductor with an inductance of L henries (H).

Ohm’s Law gives the drop in voltage due to the resistor as RI. The voltage drop due to the inductor is L(dI/dt).

One of Kirchhoff’s laws says that the sum of the voltage drops is equal to the supplied voltage E(t).

Application to Electric Circuits

Thus we have

which is a first-order linear differential equation.

The solution gives the current I at time t.

Example 4

Suppose that in the simple circuit of Figure 4 the resistance is 12 Ω and the inductance is 4 H. If a battery gives a

constant voltage of 60 V and the switch is closed when t = 0 so the current starts with I(0) = 0, find (a) I(t), (b) the current after 1 second, and (c) the limiting value of the current.

Example 4 – Solution

(a) If we put L = 4, R = 12, and E(t) = 60 in Equation 7, we obtain the initial-value problem

or

Multiplying by the integrating factor we get

Example 4 – Solution

Since I(0) = 0, we have 5 + C = 0, so C = –5 and

cont’d

Example 4 – Solution

(b) After 1 second the current is

I(1) = 5(1 – e^–3) ≈ 4.75 A

(c) The limiting value of the current is given by

cont’d