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.
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
Continues below ⇩
`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:
ay'' + by' + cy = 0
Associated auxiliary equation:
am2 + bm + c = 0
|Nature of roots||Condition||General Solution|
|1. Real and distinct roots,
|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 + jω
m2 = α − jω
|b2 − 4ac < 0||`y = e^(alphax)(A cos \omega x + B sin \omega x)`|
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.
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`:
Graph of a possible current at time `t`, given by `i=0.1e^(-10t)+2e^(-50t)`.
Solve the following equation in which s is the displacement of an object at time t.
`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).
Now to find the values of the constants:
`s(0)=1` implies `A=1`
So we can write
`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)`.
Solve the equation
This time the auxiliary equation is:
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)`
`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`:
Graph of a typical solution given by `y=e^x(0.5 cos sqrt3 x+0.2 sin sqrt3 x)`.