Lecture Notes in Differential Equations - Bruce E. Shapiro
Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro
26 LESSON 4. LINEAR EQUATIONS where A and B do not depend on x. Hence something is linear in y if it can be written as Ay + B (4.6) if A and B do not depend on y. Thus for a differential equation to be linear in y it must have the form dy dt = Ay + B (4.7) where neither A nor B depends on y. However, both A and B are allowed to depend on t. To emphasize this we write the terms A and B as A(t) and B(t), so that the linear equation becomes dy dt = A(t)y + B(t) (4.8) For convenience of finding solutions (its not clear now why this is convenient, but trust me) we bring the term in A(t)y to the left hand side of the equation: dy − A(t)y = B(t) (4.9) dt To be consistent with most textbooks on differential equations we will relabel A(t) = −p(t) and B(t) = q(t), for some function p(t) and q(t), and this gives us dy + p(t)y = q(t) (4.10) dt which we will refer to as the standard form of the linear ODE. Sometimes for convenience we will omit the t (but the dependence will be implied) on the p and q and write the derivative with a prime, as y ′ + py = q (4.11) There is a general technique that will always work to solve a first order linear ODE. We will derive the method constructively in the following paragraph and then give several examples of its use. The idea is look for some function µ(t) that we can multiply both sides of the equation: µ × (y ′ + py) = µ × q (4.12) or µy ′ + µpy = µq (4.13)
27 So far any function µ will work, but not any function will help. We want to find an particular function µ such that the left hand side of the equation becomes d(µy) = µy ′ + µpy = µq (4.14) dt The reason for looking for this kind of µ is that if we can find µ then d (µy) = µq (4.15) dt Multiplying both sides by dt and integrating gives ∫ ∫ d dt (µ(t)y)dt = µ(t)q(t)dt (4.16) Since the integral of an exact derivative is the function itself, ∫ µ(t)y = µ(t)q(t)dt + C (4.17) hence dividing by µ, we find that if we can find µ to satisfy equation 4.14 then the general solution of equation 4.10 is y = 1 [∫ ] µ(t)q(t)dt + C (4.18) µ(t) So now we need to figure out what function µ(t) will work. product rule for derivatives From the d dt (µy) = µy′ + µ ′ y (4.19) Comparing equations 4.14 and 4.19, µy ′ + µ ′ y = µy ′ + µpy (4.20) µ ′ y = µpy (4.21) µ ′ = µp (4.22) Writing µ ′ = dµ/dt we find that we can rearrange and integrate both sides of the equation: ∫ dµ = µp (4.23) dt dµ = pdt (4.24) µ ∫ 1 µ dµ = pdt (4.25) ∫ ln µ = pdt + C (4.26)
- Page 1 and 2: Lecture Notes in Differential Equat
- Page 3 and 4: Contents Front Cover . . . . . . .
- Page 5 and 6: CONTENTS v Dedicated to the hundred
- Page 7 and 8: Preface These lecture notes on diff
- Page 9 and 10: Lesson 1 Basic Concepts A different
- Page 11 and 12: 3 Definition 1.2 (Solution, ODE). A
- Page 13 and 14: 5 1.22 is restricted to being a pos
- Page 15 and 16: 7 Figure 1.1 illustrates what this
- Page 17 and 18: 9 We will study linear equations in
- Page 19 and 20: Lesson 2 A Geometric View One way t
- Page 21 and 22: 13 We can extend this geometric int
- Page 23 and 24: 15 that since the slope of the solu
- Page 25 and 26: Lesson 3 Separable Equations An ODE
- Page 27 and 28: 19 Since it is not possible to solv
- Page 29 and 30: 21 where M(t) = −a(t) and N(y) =
- Page 31 and 32: 23 Example 3.10. Find a general sol
- Page 33: Lesson 4 Linear Equations Recall th
- Page 37 and 38: 29 Example 4.1. Solve the different
- Page 39 and 40: 31 Since p(t) = 1 (the coefficient
- Page 41 and 42: 33 Multiplying equation 4.68 by µ
- Page 43 and 44: 35 Substituting y = t = 0 in this i
- Page 45 and 46: 37 ∫ t t 0 Evaluating the integra
- Page 47 and 48: Lesson 5 Bernoulli Equations The Be
- Page 49 and 50: 41 This is a Bernoulli equation wit
- Page 51 and 52: Lesson 6 Exponential Relaxation One
- Page 53 and 54: 45 Exponential Runaway First we con
- Page 55 and 56: 47 Figure 6.2: Illustration of the
- Page 57 and 58: 49 This is identical to with Theref
- Page 59 and 60: 51 this becomes a first-order ODE i
- Page 61 and 62: Lesson 7 Autonomous Differential Eq
- Page 63 and 64: 55 Figure 7.1: A plot of the right-
- Page 65 and 66: 57 Figure 7.2: Solutions of the log
- Page 67 and 68: 59 Figure 7.4: Solutions of the thr
- Page 69 and 70: Lesson 8 Homogeneous Equations Defi
- Page 71 and 72: 63 where z = y/t, the differential
- Page 73 and 74: Lesson 9 Exact Equations We can re-
- Page 75 and 76: 67 Now compare equation (9.2) with
- Page 77 and 78: 69 Hence dg dy = 0 =⇒ g = C′ (9
- Page 79 and 80: 71 From the first of equations (9.5
- Page 81 and 82: 73 Differentiating equations (9.81)
- Page 83 and 84: 75 This has the form Mdt + Ndy = 0
26 LESSON 4. LINEAR EQUATIONS<br />
where A and B do not depend on x. Hence someth<strong>in</strong>g is l<strong>in</strong>ear <strong>in</strong> y if it<br />
can be written as<br />
Ay + B (4.6)<br />
if A and B do not depend on y. Thus for a differential equation to be l<strong>in</strong>ear<br />
<strong>in</strong> y it must have the form<br />
dy<br />
dt<br />
= Ay + B (4.7)<br />
where neither A nor B depends on y. However, both A and B are allowed<br />
to depend on t. To emphasize this we write the terms A and B as A(t)<br />
and B(t), so that the l<strong>in</strong>ear equation becomes<br />
dy<br />
dt<br />
= A(t)y + B(t) (4.8)<br />
For convenience of f<strong>in</strong>d<strong>in</strong>g solutions (its not clear now why this is convenient,<br />
but trust me) we br<strong>in</strong>g the term <strong>in</strong> A(t)y to the left hand side of the<br />
equation:<br />
dy<br />
− A(t)y = B(t) (4.9)<br />
dt<br />
To be consistent with most textbooks on differential equations we will relabel<br />
A(t) = −p(t) and B(t) = q(t), for some function p(t) and q(t), and<br />
this gives us<br />
dy<br />
+ p(t)y = q(t) (4.10)<br />
dt<br />
which we will refer to as the standard form of the l<strong>in</strong>ear ODE. Sometimes<br />
for convenience we will omit the t (but the dependence will be implied) on<br />
the p and q and write the derivative with a prime, as<br />
y ′ + py = q (4.11)<br />
There is a general technique that will always work to solve a first order<br />
l<strong>in</strong>ear ODE. We will derive the method constructively <strong>in</strong> the follow<strong>in</strong>g paragraph<br />
and then give several examples of its use. The idea is look for some<br />
function µ(t) that we can multiply both sides of the equation:<br />
µ × (y ′ + py) = µ × q (4.12)<br />
or<br />
µy ′ + µpy = µq (4.13)