MATH 3271 - PARTIAL DIFFERENTIAL EQUATIONS

ODE SUMMARY SHEET

Note that this is not meant to be an exhaustive list of everything you shouldrecall from your ODEs course. Rather, it is meant to be a helpful summary tohelp you recall some important facts.

1. First-Order ODEs

Consider a general first-order linear ordinary differential equation:

(1) y′(t) + p(t)y = g(t).

We define an integrating factor as

(2) µ(t) = exp

∫p(t)dt.

Then, the solution to this ODE is given by

(3) y(t) =1

µ(t)

[∫ t

t0

µ(s)g(s)ds+ c

], c ∈ R.

Consider now a more general first-order ODE given as

(4)dy

dx= f(x, y).

If this is nonlinear, then there is no universal method for solving the equation. Rather, wewill consider a subclass of first-order equations that can be solved by simply integrating.We will re-write this ODE in the form

(5) M(x, y) +N(x, y)dy

dx= 0.

For example, we typically set M(x, y) = −f(x, y), and N(x, y) = 1, but there are clearlyother possibilities. If we can perform this separation such that M is a function of x onlyand N is a function of y only, then this equation takes the simple form

(6) M(x) +N(y)dy

dx= 0,

or

(7) M(x)dx+N(y)dy = 0.

So, this separable equation can be solved by simply integrating the functions M and N .

Date: September 14, 2015 - Ikjyot Singh Kohli.

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2. Second-Order ODEs

Consider a second-order linear homogeneous ODE of the form

(8) ay′′ + by′ + cy = 0.

Since this is a second-order ODE, we require two independent solutions of this ODE. Recallthat related to this ODE is the corresponding characteristic polynomial

(9) ar2 + br + c = 0.

The type of roots r1, r2 of this equation determine the general solution to this second-orderODE.

In particular, we have that if r1, r2 ∈ R, r1 6= r2, then

(10) y = c1er1t + c2e

r2t, c1, c2 ∈ R .

If r1, r2 are conjugate complex numbers such that

(11) r1 = λ+ iµ, r2 = λ− iµ,then the solutions to the ODE are given by

(12) y = c1 exp [(λ+ iµ) t] + c2 exp [(λ− iµ) t] , c1, c2 ∈ R.Note that, this is not the typical form given for solutions to the second-order ODE in thiscase, since the solutions as presented are complex-valued. Rather, we would like real-valuedsolutions. We accomplish this as follows. Recall from the principle of superposition thatif y1 and y2 are solutions of the second-order ODE under consideration, then any linearcombination c1y1 + c2y2 is also a solution. Therefore, let us denote

(13) y1 = exp [(λ+ iµ) t] , y2 = exp [(λ− iµ) t] .

We first compute

(14) y1 + y2 = eλt (cosµt+ i sinµt) + eλt (cosµt− i sinµt) = 2eλt cosµt.

and

(15) y1 − y2 = eλt (cosµt+ i sinµt)− eλt (cosµt− i sinµt) = 2ieλt sinµt.

Note that the factors of 2 and 2i are simply just constant multipliers, and therefore can beneglected. We have therefore obtained real-valued solutions to our second-order ODE inthis case, given by

(16) y = c1eλt cosµt+ c2e

λt sinµt, c1, c2 ∈ R .

For our final case, consider when r1 = r2 ∈ R, that is, we only have one distinct root ofthe characteristic polynomial. In this case, we have from solving

(17) ar2 + br + c = 0,

that

(18) r1 = r2 = − b

2a.

MATH 3271 - PARTIAL DIFFERENTIAL EQUATIONS 3

Therefore, we only have one solution

(19) y1(t) = e−bt2a ,

which is a problem, since we require two linearly independent solutions. We therefore needa second solution that is not a multiple of y1. Let us assume a second solution exists ofthe form

(20) y = v(t)y1(t) = v(t)e−bt/2a.

We then compute

(21) y′ = v′(t)e−bt/2a − b

2av(t)e−bt/2a,

and

(22) y′′ = v′′(t)e−bt/2a − b

av′(t)e−bt/2a +

b2

4a2v(t)e−bt/2a.

Substituting these expressions into Eq. (8), we obtain the remarkable result

(23) v′′(t) = 0.

This implies that

(24) v(t) = c1t+ c2, c1, c2 ∈ R.

Therefore, from Eq. (20), we have that

(25) y = c1te−bt/2a + c2e

−bt/2a, c1, c2 ∈ R ,

and this is the required solution.

3. Euler Equations

Later on in this course, we will be required to know how to solve the so-called Eulerequation:

(26) x2y′′ + αxy′ + βy = 0, α, β ∈ R.

To solve the Euler equation, we assume a solution of the form

(27) y = xr.

Substituting this into Eq. (26), we obtain

(28) xr [r(r − 1) + αr + β] .

Therefore, if r is a root of the quadratic equation

(29) F (r) = r(r − 1) + αr + β = 0,

then y = xr is indeed a solution of Eq. (26). Further, it can be shown that if the roots ofEq. (29) are real and different then,

(30) y = c1|x|r1 + c2|x|r2 , c1, c2 ∈ R.

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If the roots are real and equal, then

(31) y = (c1 + c2 ln |x|) |x|r2 , c1, c2 ∈ R.If the roots are complex conjugates, then

(32) y = |x|λ [c1 cos(µ ln |x|) + c2 sin(µ ln |x|)] , c1, c2 ∈ R, r1, r2 = λ± iµ.