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Week 2

In week 2 we began to look at homogenous second order linear differential equations (with constant coefficients). These are differential equations of the form

a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0,               (*)

where a,\,b,\,c are real constants and the solution is y=f(x). We showed that to solve these equations we look at the auxiliary equation

ar^2+br+c=0.

Fact 1

In general, if the roots of of this equation are \alpha and \beta then the solution to  (*) is given by

y(x)=Ae^{\alpha x}+Be^{\beta x}

where A and B are real constants.

The real constants are to be determined by two boundary/initial conditions of the form y(x_i)=y_i and \left.\frac{dy}{dx}\right|_{x=x_j}=y'_j.

There were possible complications however. If the roots are real and equal, say both \alpha, then the naive application of the above method yields:

y(x)=Ae^{\alpha x}+Be^{\alpha x}=(A+B)e^{\alpha x}=Ce^{\alpha ^x},

which at the end of the day is just ‘constant’ \times e^{\alpha x}.

A good question at this point is how do we know there are more solutions (as we are going to say below)? Before we even answer this we should note that the y(x) written here cannot satisfy a boundary condition of the form y(0)=0 unless C=0 and this isn’t very interesting at all… i.e. y(x)=0e^{\alpha x}=0.

We can do even better and design an actual damped harmonic oscillator/door closer such that the roots of the auxiliary equation are repeated. Now open the door to a width of 1 m and release from rest: x'(0)=0 — the initial speed is zero. Clearly something is missing and we can say this without going to deeply into the mathematics.

If we do go deeper into the mathematics we can go away and prove that the solution space of the linear operator:

a\frac{d^2}{dx^2}+b\frac{d}{dx}+cI

is two dimensional (briefly, needs two independent components) so there must be another solution. As it turns out, and we did the calculation in class, Bxe^{\alpha x} will be a second solution.

Fact 2

If the roots of the auxiliary equation to (*) are repeated, then the general solution is given by


y(x)=Ae^{\alpha x}+Bxe^{\alpha x}.

A second complication that can occur is if the roots are complex. The Conjugate Root Theorem says that the roots of the auxiliary equation will be of the form

p\pm iq

whenever the coefficients a,\,b,\,c are real (which they will be). In this case we could write the solution as

y(x)=Ae^{p+iq}+Be^{p-iq},

for some possibly complex constants A and B.

However it can be shown that this solution can be rewritten. This yields another fact:

Fact 3

If the roots of the auxiliary equation to (*) are complex, then the general solution is given by

y(x)=e^{px}(C_1\cos qx +C_2\sin qx),

for real constants C_1, and C_2.

Week 3

In week 3 we asked what happens when the right hand side of (*) is not zero but rather some non-zero function of x:

a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x)      (**)

This is then a non-homogeneous second order linear differential equation. We were able to prove the following fact:

N-HSOLDE Fact

The general solution of (**) is given by

y_G(x)=y_H(x)+y_P(x),

where y_H(x) is the solution to the associated homogeneous differential equation (*) and y_P(x) is a particular (i.e. any) solution of (**).

The trick is to find a particular solution based on what f(x). In general;

If f(x)=C, try y_{\text{trial}}=k. If f(x)=C_1x+C_2, y_T=k_1x+k_0. If f(x)=Cx^2, y_T=k_2x^2+k_1x+k_0. If f(x)=Cx^n, y_T=k_nx^n+\cdots+k_1x+k_0.

If f(x)=Ce^{qx}, y_T=ke^{qx} and y_T=C\sin qx or C\cos qx, y_T=k_1\cos qx+k_2\sin qx. If it happens that y_T does not satisfy (**) then it probably happens that y_T is in fact a solution to (*). Try y_{T,2}=xy_T in this case.

Tutorials

Tutorials Tuesday 6 – 8 pm start this week. If you have any questions this will be the time to ask. After that we will have tutorials in weeks 4, 5, 6, 8, 10, 12.

Test 1

As it stands the test will be on in week 6 although I might be moving this to week 7. I will have a sample test for ye very soon.

Notes

The notes on second order linear differential equations that I handed out. Note that you should be attempting the exercises.

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