Roots of Homogeneous Second Order ODE's and the Nature their Solutions

in #steemstem6 years ago

Continuing our second-order differential equations journey, let's have a look at some possible solutions to the homogeneous second order linear ODE...

p1.png

Now, to recap, the characteristic equation corresponding to equation (1) is...

p2.png

And the roots of equation (2) are...

p3.png

Let's call the term inside the root sign Δ. Therefore...

p4.png

The roots of the characteristic equation determines the nature of the solution, and types of roots we can obtain from the characteristic equation are determined by Δ. They are...

Case I > 0 2 real and distinct roots
Case II = 0 1 real double root
Case III < 0 2 complex conjugate roots

Let's have a look at an example of each case...

Case I: 2 real and distinct roots when Δ > 0

For Case I, the corresponding general solution is relatively straightforward as we simply get...

p5.png

So let's do an example, and find the particular solution to the equation...

p6.png

with initial conditions...

p7.png

Firstly, the characteristic equation corresponding to (4) is...

p8.png

And the roots to equation (5) are...

p9.png

And thus the general solution to (4) is...

p10.png

Now, applying the initial conditions, we have...

p11.png

Solving (6) & (7) simultaneously, we get A = 6 and B = 4. And thus the particular solution is...

p12.png

A graph of the solution is shown in Figure 1 below....

p15.png
Figure 1. The particular solution to equation (4).

Case II: 1 real double root when Δ = 0

By the quadratic formula, when the discriminant Δ = 0, there is only 1 root...

p16.png

So the solution is not simply...

p17.png

...because we've only found 1 solution. To form a basis for 2nd Order ODE's, we need 2 linearly independent solutions.

To find the other solution, we need to use the reduction of order method discussed in the previous post.

By equation (4) of the previous post, we have...

p18.png

and...

p19.png

Therefore, the second solution that forms the basis is p20.png. And thus the general solution is...

p21.png

Let's do an example that only has 1 real root. Take the equation...

p22.png

with initial conditions...

p23.png

The characteristic equation of (9) is...

p24.png

And the only root is...

p25.png

And thus the general solution is...

p26.png

Now, applying the Initial Conditions, we have...

p27.png

Therefore, the particular solution is...

p28.png

Graphically, the solution is shown below in Figure 2...

p31.png
Figure 2. The particular solution to equation (9)

Case III: Complex Conjugate Roots when Δ < 0

When the discriminant is negative, the roots of the characteristic equation are...

p32.png

Thus we have 2 complex conjugate roots, where...

p33.png

Thus the general solution is...

p34.png

But how do we decipher the exponential functions of an imaginary number? Thankfully, we have the ingenious Euler's formula, where...

p35.png

So substituting these into the general solution, we get...

p36.png

Now, here's another neat trick. If A and B are complex conjugate pairs, that is, let...

p37.png

...then the general solution becomes...

p38.png

...where C and D are real, undetermined coefficients.

Now let's do one last example that has 2 complex conjugate roots...

p39.png

...with initial conditions...

p50.png

The characteristic equation of (12) is...

p40.png

Thus the roots are...

p41.png

And the general solution is...

p42.png

Applying the first initial condition to (13), we have...

p43.png

For the second initial conditon, we need the first derivative of (13), which is...

p44.png

So applying the second initial condition, we have...

p45.png

And thus, after all that the particular solution is...

p46.png

Finally, Figure 3 below illustrates this particular solution...

p49.png
Figure 3. The particular solution of equation (12)


Credits:

All equations in this tutorial were created with QuickLatex

All graphs were created with www.desmos.com/calculator

All example problems are my worked solutions from the textbook Advanced Engineering Mathemathics, 10th Edition, Kreyszig


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