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7. Second Order Homogeneous Linear DEs With Constant Coefficients

The general form of the second order differential equation with constant coefficients is


where a, b, c are constants with a > 0 and Q(x) is a function of x only.

Homogeneous Equation

In this section, most of our examples are homogeneous 2nd order linear DEs (that is, with Q(x) = 0):


where a, b, c are constants.

Method of Solution

The equation

`am^2 + bm + c = 0 `

is called the Auxiliary Equation (A.E.)

The general solution of the differential equation depends on the solution of the A.E. To find the general solution, we must determine the roots of the A.E. The roots of the A.E. are given by the well-known quadratic formula:



Differential Equation:

ay'' + by' + cy = 0

Associated auxiliary equation:

am2 + bm + c = 0

Nature of roots Condition General Solution
1. Real and distinct roots,
m1, m2
b2 − 4ac > 0 `y = Ae^(m_1x)+Be^(m_2x)`
2. Real and equal roots, m b2 − 4ac = 0 `y = e^(mx)(A + Bx)`

3. Complex roots

m1 = α +

m2 = α

b2 − 4ac < 0 `y = e^(alphax)(A cos \omega x + B sin \omega x)`

Example 1

The current i flowing through a circuit is given by the equation


Solve for the current i at time t > 0.


The auxiliary equation arising from the given differential equations is:

A.E.: `m^2+60m+500` `=(m+10)(m+50)` `=0`

So `m_1=-10` and `m_2=-50`.

We have 2 distinct real roots, so we need to use the first solution from the table above (y = Aem1x + Bem2x), but we use i instead of y, and t instead of x.

So `i=Ae^(-10t)+Be^(-50t)`

We could have written this as:




since we have 2 constants of integration. We would be able to find these constants if we were given some initial conditions.

Here's a typical solution graph for Example 1 with arbitrary values `A=0.1` and `B=2`:

Example 2

Solve the following equation in which s is the displacement of an object at time t.


given that

`s = 1`, `(ds)/(dt)=3` when `t=0.`

(That is, the object's position is 1 unit and its velocity is 3 units at the beginning of the motion.)


The auxiliary equation for our differential equation is:

A.E. `m^2-4m+4` `=(m-2)^2` `=0`

In this case, we have:

`m=2` (repeated root)

We need to use the second form from the table above (`y = e^(mx)(A + Bx)`), and once again use the correct variables (t and s, instead of x and y).

So `s(t)=(A+Bt)e^(2t)`

Now to find the values of the constants:

`s(0)=1` implies `A=1`

So we can write


Differentiating gives:



`s'(0)=3 ` implies `2+B=3`

This gives us `B=1`



The graph of our solution is as follows:

Graph of displacement `s(t)=(1+t)e^(2t)`.

Example 3

Solve the equation



This time the auxiliary equation is:

A.E. `m^2-2m+4=0`

Solving for m, we find that the solutions are a complex conjugate pair:

`m_1=1-jsqrt3` and `m_2=1+jsqrt3`

The solution for our DE, using the 3rd type from the table above:

`y = e^(alphax)(A\ cos\ \omega x + B\ sin\ \omega x)`

we get:

`y(x)=` `e^x(A\ cossqrt3 x+B\ sinsqrt3 x)`

Here's a typical solution graph for Example 3 with arbitrary values `A=0.5` and `B=0.2`:

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