Differential Equations Class Notes

Differential Equations Class Notes Dan Wysocki Spring 213 Contents 1 Introduction 2 2 Classification of Differential Equations 6 2.1 Linear vs. Non-Linear.................................. 7 2.2 Seperable Equations................................... 8 2.3 Exact Differential Equations............................... 1 2.4 Integrating Factor..................................... 15 2.5 Modeling with First Order Equations.......................... 21 2.6 Bernoulli s Equation................................... 24 2.7 Homogenous Equations.................................. 26 2.8 Second Order Linear Equations............................. 26 3 Transformations 33 1

3.1 Laplace Transform.................................... 33 3.2 Inverse Laplace Transform................................ 35 3.3 Unit Step-function.................................... 42 3.4 Gamma Function..................................... 47 4 Systems of Differential Equations 52 5 Final Examples 61 2

1 Introduction Differential equations are equations that have one or more derivatives or differentials. 1 x dy dx = 1 y First order (1.1) 2x + y y = 5 Second order (1.2) (1.3) y = dy dx 4 d2 y + xy = sin x (1.4) dx2 What is a solution to a Differential Equation? Show that y(x) = x 2 is a solution. x dy dx = 2y (1.5) dy dx = 2x x (2x) = 2(x 2 ) 2x 2 = 2x 2 For (1 + xe xy ) dy dx + 1 + yexy =, (1.6) show that x + y + e xy = is a solution. 3

Implicit Solution d dx (x + y + exy ) = d dx () 1 + dy ( dx + exy y + x dy ) = dx 1 + dy dx + y exy xy dy + x e dx = dy dx (1 + x exy ) + 1 + y e xy = For dy dx = 2x, (1.7) show that the following are solutions 1. y(x) = x 2 d dx y(x) = d dx x2 dy dx = 2x 2. y(x) = x 2 + 1 d dx y(x) = d dx x2 + 1 dy dx = 2x 4

3. y(x) = x 2 + C d dx y(x) = d dx x2 + C dy dx = 2x Show that p(t) = 9 + C e t /2 is a solution. dp dt = 1 p 45 (1.8) 2 dp dt = + C et /2 1 2 = C 2 et /2 LHS = C 2 et /2 RHS = 1 2 = C 2 et /2 ( 9 + C e t/2 ) 45 What if p() = 85 (initial condition) 85 = 9 + C e /2 85 = 9 + C C = 5 We now have an Initial Value Problem (IVP). dp = 1p 45 dt 2 p() = 85 5

Solution is p(t) = 9 5 e t /2 dy dx = f(x, y), dy dx is the slope. Example: Consider a mouse population that reproduces at a rate proportional to the current population, with a rate constant equal to 1 /2 mice /month assuming no owls present. When owls are present, they eat the mice. Suppose that the owls eat on average 15 mice per day. Assuming there are 3 days in a month, write a differential equation to describe the above. p = mouse population, t = time in months dp dt p dp dt = kp = 1 2 p 45, (45 is the mice eaten per month) When p = 9, dp =. This is the equilibrium solution. dt When p > 9, dp dt > When p < 9, dp dt < 6

Next time: Classifications of Diff Eqns & Seperable Diff Eqns. 2 Classification of Differential Equations Ordinary vs. Partial Differential Equation (ODE vs. PDE) based on number of independent variables (dedpendent variable) y = f(x (independent variable)) If there is one independent variable ODE If there is more than one independent variable PDE Example 2.1 dp dt = 1 p 45 p(t) one independant variable (2.1) 2 Example 2.2 2 v(x, t) x 2 = v(x, t) t PDE (Heat equation) (2.2) 7

Example 2.3 2 v(x, t) x 2 = 2 v(x, t) t 2 PDE (Wave equation) (2.3) Based on number of unknowns system of equations dx = 4x + 3t dt dy = 5x 2y dt Order of a differential equation is the highest derivative that appears in the equation. Equation 1 y + 3y = 1 st order linear Equation 2 y + 3y 2t = 2 nd order linear d Equation 3 y dt d2 y 4 dt + 1 = 2 e2t 4 th order linear Equation 4 u xx + u t = 2 nd order linear Equation 5 u xx + u u yy = sin t 2 nd order non-linear Equation 6 u xx + sin(u)u yy = cos t 2 nd order non-linear Equation 7 u xx + u yy + sin(u) = 2 nd order nonlinear Equation 8 u xx + u yy + sin(t) = 2 nd order linear Equation 9 u xx + u yy + u = 2 nd order linear Table 1: Examples of different order and linear vs. non-linear differential equations. 2.1 Linear vs. Non-Linear A linear ODE is of the form: a n (t)y (n) + a n 1 (t)y (n 1) +... + a 1 (t)y + a (t)y = g(t) (2.4) This is similar to a polynomial, which take the form a n x n + a n 1 x n 1 +... + a 1 x + a = f(x) (2.5) 8

But instead of it being the independent variables raised to decreasing powers, it s the derivative of the independent variables with decreasing orders of the derivatives. 2.2 Seperable Equations Consider a first order ODE dy dx = f(x, y)... (2.6) If we can express Equation 2.6 in the form of M(x) dx + N(y) dy = (2.7) then Equation 2.6 is called a seperable equation. Solve dy dx = x2 1 y 2 (2.8) (1 y 2 ) dy = N(y) dy = x 2 dx M(x) dx y y3 3 = x3 + C (Implicit Solution) 3 Solve dy dx = 3x2 4x + 2 2(y 1) (2.9) 9

2(y 1) dy = 3x 2 4x + 2 dx y 2 2y = x 3 2x 2 + 2x + C Complete the square by adding 1 to both sides (note that C is still a constant, so the 1 disappears) (y 1) 2 = x 3 + 2x 2 + 2x + C y 1 = ± x 3 + 2x 2 + 2x + C y = 1 ± x 3 + 2x 2 + 2x + C (Explicit Solution) If, also, y() = 1, solve for y 1 = 1 ± 3 + 2 + + C = 1 ± C 2 = ± C C = ( 2) 2 = 4 y = 1 ± x 3 + 2x 2 + 2x + 4 1 = 1 ± 4 = 1 ±2 y = 1 x 3 + 2x 2 + 2x + 4 How about y() = 3, C = 4? 3 = 1 ± + 4 = 1 ±2 y = 1 + x 3 + 2x 2 + 2x + 4 1

Solve y y() = 1 = 4x x3 4+y 3 (2.1) dy 4x x3 = dx 4 + y 3 4 + y 3 dy = 4x x 3 dx 4y + 1 4 y4 = 2x 2 1 4 x4 + C 4(1) + 1 4 14 = + C 4 + 1 4 = C C = 17 4 4y + 1 4 y4 = 2x 2 1 4 x4 + 17 4 2.3 Exact Differential Equations Assume we have a solution Ψ(x, y) = C (2.11) Then we will find what this solution is. Theorem 2.4 M(x, y) + N(x, y) dy dx = is called exact if M(x, y) y = N(x, y). x For Equation 2.11, Ψ x Ψ x + Ψ dy =. (2.12) y dx = M(x, y), Ψ y = N(x, y) (2.13) 11

Example 2.5 2x + y 2 + 2xy dy dx = M y 2x + y 2 = M(x, y) 2xy = N(x, y) N = 2y, x = 2y Ψ (2x x = ) + y 2 dx = x 2 + xy 2 + c 1 Ψ y = 2xy dy = xy 2 + c 2 Ψ(x, y) = x 2 + xy 2 + h(y) Ψ(x, y) = xy 2 + h(x) Ψ y = 2xy + h (y) h (y) = h (y)dy = dy h(y) = c 3 x 2 + xy 2 + c 3 = c 4 x 2 + xy 2 = C Example 2.6 2xy + ( x 2 + 1 ) dy dx = 2xy = M(x, y) x 2 + 1 = N(x, y) 12

First we must see if they are exact M y = 2x N x = 2x M y = N x x 2 y + y = C Ψ(x, y) = 2xy x Ψ(x, y) dx = 2xy dx x Ψ(x, y) = x 2 y + h(y) Now take the derivative with respect to y and compare it to the known Ψ y Ψ y = x2 + h (y) x 2 + h (y) = x 2 + 1 h (y) dy = 1 dy to find h(y) h(y) = y + c 1 Ψ(x, y) = x 2 y + y + c 1 x 2 y + y = C Example 2.7 ( x 2 + y ) + (x + cos(y)) dy dx = M(x, y) = x 2 + y N(x, y) = x + cos(y) M x = 2x N x = 13

Example 2.8 2xy + (x 2 + 1) dy dx = M y = 2x, N x = 2x 14

Example 2.9 x 2 dy dx = x3 2xy y(1) = 3 M(x, y) = x 3 + 2xy N(x, y) = x 2 M y = 2x N x = 2x It is exact, so there must be a solution of the form: Ψ(x, y) = C Ψ x + Ψ dy y dx = Ψ x = x3 + 2xy Ψ y = x2 Integrate: Ψ y = x 2 dy Ψ(x, y) = x 2 y + h(x) Ψ x = 2xy + h (x) x 3 + 2xy = 2xy + h (x) h (x) = x 3 h(x) = x 3 dx = x4 4 + C Ψ(x, y) = x 2 y x4 4 + C x 2 y x4 4 = C 15

y(1) = 3 1 2 3 14 4 = C 3 1 4 = C C = 11 /4 x 2 y x4 4 = 11 /4 Example 2.1 (x 2 + y) + (x sin(y)) dy dx = M = (x 2 + y), N = (x sin(y)) It is exact M y = 1 N x = 1 2.4 Integrating Factor Example 2.11 x 2 + y x + x sin(y) dy x M = x2 + y x It is not exact dx =, N = x sin(y) x M y = 1 /x N x = sin(y) x 2 Both examples have the same solution, but one is exact and the other is not. Example 2.11 is simply Example 2.1 divided by x. 16

Key term (x) has been divided out! Key term Integrating factor. y + p(x)y = g(x) (2.14) The Integrating factor is µ(x) = e p(x) dx µ (x) = e p(x) dx d dx p(x) dx µ (x) = e p(x) dx p(x) µ (x) = µ(x) p(x) µ(x) y + µ(x) p(x) y = µ(x) g(x) µ(x) y + µ (x) y = µ(x) g(x) d (µ(x) y) = µ(x) g(x) dx dx µ(x) y = µ(x) g(x) dx y = 1 µ(x) g(x) dx µ(x) Example 2.12 Solve y + 2y = e x p(x) = 2, g(x) = e x µ(x) = e 2 dx = e 2x y = 1 e 2x e x dx e 2x = e 2x e 3x dx ( ) 1 = e 2x 3 e3x + c = 1 3 ex + c e 2x 17

Example 2.13 dy dx = 4 2y x dy dx + 2y x = 4 p(x) = 2 x, g(x) = 4 µ(x) = e p(x) dx = e 2 ln x ln(a b ) = b ln(a) ln x2 = e = x 2 y = 1 4 x 2 x 2 ( ) 4 = x 2 3 x3 + c y(x) = 4 3 x + c x 2 Example 2.14 dy + 2xy = 1, dx y(1) = 2 p(x) = 2x, g(x) = 1 µ(x) = e p(x) dx = e x2 y = 1 e x2 e x2 dx You can t integrate e x2 18

Example 2.15 x 2 y 3 + x(1 + y 2 )y = M y = 3x 2 y 2, N x = 1 + y 2 M y N x Not exact µ(x, y) = 1 xy 3 x 2 y 3 xy + x(1 + y2 ) y = 3 xy ( 3 1 x + y + 1 ) y = 3 y M y =, N x = M y = N x Exact! M dx = x dx = x2 2 + C (y N dy = 3 + y 1) dy = 1 + ln y + C 2y2 2 1 + ln y = C 2y2 x 2 Theorem 2.16 If N x M y M M + Ny = = Q(y), then has an integrating factor µ(y) = e Q(y) dy 19

µm + µny = e Q(y) dy M + e Q(y) dy Ny = M y = e Q(y) dy M Ñ x = e Q(y) dy N M y = Ñx M y = y ( e ) Q(y) dy M = e Q(y) dy Q(y)M + e Q(y) dy M y Ñ x = ( e ) Q(y) dy N x = e Q(y) dy N x N x M y = MQ(y) N x = M y + MQ(y) Ñ x = e Q(y) dy (M y + MQ(y)) M y = Ñx, Exact! = e Q(y) dy M y + MQ(y)e Q(y) dy = M y 2

Example 2.17 y + ( 2xy e 2y) y = N(x, y) = 2xy e 2y M(x, y) = y Using Theorem 2.16: N x = 2y, M y = 1 N x M y M = 2y 1 y Q(y) = 2 1 y µ(y) = e 2 1 y dy 2y ln y = e = e 2y ln y e = e 2y e ln 1 y = 1 y e2y 1 y e2y [ y + ( 2xy e 2y) y ] = e 2y + 1 ( y e2y 2xy e 2y) y = ( e 2y + 2xe 2y 1 ) y = y = 2 1 y M = e 2y ( N = 2xe 2y 1 ) y M y = 2e 2y N x = 2e 2y M y = N x Exact! M dx = xe 2y N dy = xe 2y ln y xe 2y ln y = C 21

2.5 Modeling with First Order Equations Example 2.18 Mixing Problem At time t =, a tank contains Q lb of salt dissolved in 1 gal of water. Assume that water containing 1 /4 lb of salt/gal is entering the tank at a rate of r gal/min and that the well-stirred mixture is draining from the tank at the same rate. Set up the initial value problem that describes this flow process. Find the amount of salt Q(t) in the tank at any time, and also find the limiting amount Q L that is present after a very long time. If r = 3 and Q = 2Q L, find the time T after which the salt level is within 2% of Q L. Also find the flow rate that is required if the value of T is not to exceed 45 min. Let Q(t) be the amount of salt at time t dq dt = Rate in Rate out Rate in = (Amount) (Rate) Amount = 1 /4 lb /gal Rate = r gal /min Rate out: Amount : Rate : Q [lb] 1 gal r [ gal /min] 22

( dq 1 dt = 4 = 1 4 r Q 1 r dq dt + Q 1 r = 1 4 r µ(t) = e = e rt 1 ) ( lb r gal ) ( Q gal min 1 r 1 dt ) ( lb r gal ) gal min Q() = Q Example 2.19 Example 3 from Page 57 in the textbook. Let Q(t) be the amount of chemical in the pond at any time t. dq dt = Rate in Rate out = (2 + sin(2t)) g ( 5 gal ) gal year Q ( g 5 gal ) 1 gal year g = (1 + 5 sin(2t)) year Q 2 Q() = = 1 + 5 sin(2t) Q 2 g year p(t) = 1 2 g(t) = 1 + 5 sin(2t) µ(t) = e p(t) dt = e t /2 Q = e t/2 e t /2 (1 + 5 sin(2t)) dt [ ] = e t/2 1 e t /2 dt + 5 e t /2 sin(2t) dt 23

Integration by parts: f(t) = e t /2 sin(2t) dt v = e t /2 dv = sin(2t) dt dv = 1 /2 e t /2 dt v = 1 /2 cos(2t) = 1 /2 e t /2 cos(2t) + 1 /4 e t /2 cos(2t) dt g(t) = 1 /4 e t /2 cos(2t) dt v = e t /2 dv = cos(2t) dt du = 1 /2 e t /2 v = 1 /2 sin(2t) [ ] = 1 /4 1/2 e t /2 sin(2t) 1 /4 e t /2 sin(2t) dt I(t) = e t /2 sin(2t) dt v = e t /2 dv = sin(2t) dt du = 1 /2 e t /2 v = 1 /2 cos(2t) = 1 /2 e t /2 cos(2t) + 1 /4 e t /2 cos(2t) dt = 1 /2 e t /2 cos(2t) + 1 /8 e t /2 sin(2t) I 16 17 16 I = 1 /2 e t/2 cos(2t) + 1 /8 e t /2 sin(2t) + e Now we can substitute I(t) back into g(t) and then f(t) to get our final solution. Example 2.2 Page 59, Problem 3 A tank originally contains 1 gal of fresh water. Then water containing 1 /2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 1 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 1 min. 24

d Q dt Q() = = Rate in Rate out d Q dt = 1 lb min Q lb 5 min d Q dt = 1 Q 5 1 = d Q dt + Q 5 µ(t) = e 1/5 dt = e t /5 Q(t) = e t/5 e t /5 1 dt = e t/5 ( 5 e t /5 + C ) = 5 + Ce t/5 Q() = = 5 + C C = 5 Q(t) = 5 5e t/5 Q(1) = 5 5e 1/5 9.6 lb 2.6 Bernoulli s Equation Bernoulli s Equation is of the form dy dx + p(x)y = g(x)yn (2.15) This is a more general case of the form we ve used for integrating factors, where n was equal to 1 dy dx + p(x)y = g(x) y 25

Let y = v 1 1 n ; y n = v n 1 n 1 1 n 1 n 1 n dy dx = v dv 1 n dx = v dv 1 n dx dv 1 n dx + p(x)v 1 1 n { n Now divide everything by v 1 n 1 1 n = v 1 1 n n v n 1 n v = g(x)v n 1 n 1 n n 1 n dv + (1 n)p(x)v = g(x)(1 n) dx dy + p(x)y = g(x)yn dx = v 1 n 1 n = v 1 = v Example 2.21 Solve dy dx + 3y = ex y 2 This is a Bernoulli with n = 2 y = v 1 1 2 dv + (1 2)p(x)v = g(x)(1 2) dx dv p(x)v = g(x) dx dv 3v = ex dx = v 1 y = 1 v µ(x) = e 3 dx = e 3x v(x) = e 3x e 3x ( 1)e x dx = e 3x e 3x e x dx = e 3x e 2x dx ( ) = e 3x e 2x 2 + c 1 = ex 2 c 1e 3x = ex 2 + Ce3x 26

but y = 1 v = v 1 ( e x so y(x) = 2 + Ce3x ) 1 y = is a solution as well 2.7 Homogenous Equations wrote it in my notebook 2.8 Second Order Linear Equations of the form d2 y dx 2 = f(x, y, y ) or p(x)y + q(x)y + r(x)y = g(x) Example 2.22 Solve y y = or y = y. { y 1 (x) = e x, y 2 (x) = e x y(x) = c 1 y 1 (x) + c 2 y 2 (x) = c 1 e x + c 2 e x y (x) = c 1 e x c 2 e x y (x) = c 1 e x + c 2 e x = y(x) So y(x) = c 1 e x + c 2 e x is the general solution. 27

If we have ay +by +cy = where a, b, and c are constants, then y = e mx is a solution if y = me mx, y = m 2 e mx am 2 e mx + bme mx + ce mx = e mx [ am 2 + bm + c ] = So am 2 + bm + c = (characteristic equation) There are three cases for characteristic equations 1. Two distinct roots r 1 r 2 y(x) = c1e r1x + c2e r2x 2. Complex roots m = α ± βi y(x) = c 1 e αx cos(βx) + c 2 e αx sin(βx) 3. Repeated roots r 1 = r 2 = r y(x) = c 1 e rx + c 2 xe rx A second order linear equation y + p(x)y + q(x)y = g(x) is called homogenous if g(x) = and inhomogenous if g(x). If r 1 and r 2 are two distinct roots of a characteristic equation, then the general solution is given by y(x) = c 1 e r 1x + c 2 e r 2x 28

Example 2.23 Solve y + 3y 1y = e [ mx m 2 + 3m 1 ] = m 2 + 3m 1 = (m + 5)(m 2) = m = 5, m = 2 y(x) = c 1 e 5x + c 2 e 2x For an IVP you would need y(x ) = y and y (x ) = ỹ Example 2.24 I.V.P. y + 4y + 3y = y() = 1, y () = 1 m 2 + 4m + 3 = (x + 1)(x + 3) = y(x) = c 1 e x + c 2 e 3x y (x) = c 1 e x 3c 2 e 3x y() = 1 = c 1 + c 2 y () = 1 = c 1 3c 2 1 1 1 1 2 1 3 1 1 1 y(x) = 2e x e 3x 29

Example 2.25 Solve x 2 + 2x + 2 = x = 2 ± 2 2 4(1)(2) 2(1) = 1 ± i Example 2.26 Solve y + 2y + 2y = m 2 + 2m + 2 = m = 1 ± i r 1 = 1 + i r 2 = 1 i y(x) = c 1 e ( 1+i)x + c 2 e ( 1 i)x = c 1 e x+xi + c 2 e x xi = c 1 e x [cos(x) + i sin(x)] + c 2 e x [cos( x) + i sin( x)] sin(θ) = sin(θ) cos(θ) = cos( θ) = c 1 e x cos(x) + ic 1 e x sin(x) + c 2 e x cos(x) ic 2 e x sin(x) = (c 1 + c 2 )e x cos(x) + (c 1 c 2 )ie x sin(x) A = c 1 + c 2 B = i(c 1 c 2 ) y(x) = Ae x cos(x) + Be x sin(x) So, if α ± iβ are the roots, the general solution is y(x) = Ae αx cos(βx) + Be αx sin(βx) Example 2.27 The roots are: m = 2 ± 7i 3

Then the general solution must be y(x) = Ae 2x cos(7x) + Be 2x sin(7x) = e 2x [A cos(7x) + B sin(7x)] Example 2.28 y + 14y + 149y = y() = 1, y () = 8 Example 2.29 y + 4y + 13y = y() =, y () = 7 For repeated roots, r 1 = r 2, so y(x) = e r 1x = e r 2x Example 2.3 y + 4y + 4y = m 2 + 4m + 4 = (m + 2) 2 =, m = 2 y(x) = v(x)e 2x y (x) = v (x)e 2x 2v(x)e 2x = e 2x [v (x) 2v(x)] y (x) = v (x)e 2x 2v (x)e 2x + 4v(x)e 2x 2v (x)e 2x = e 2x [v (x) 4v (x) + 4v(x)] e 2x [v (x) 4v (x) + 4v(x)] + 4e 2x [v (x) 2v(x)] + 4e 2x v(x) = e 2x [v (x) 4v (x) + 4v(x) + 4v (x) 8v(x) + 4v(x)] = 31

e 2x v (x) = v = c 1 v = c 1 dx = c 1 x + c 2 v(x) = c 1 x + c 2 y(x) = (c 1 x + c 2 ) e 2x y(x) = c 2 e 2x + c 1 xe 2x y (x) = 2c 2 e 2x + c 1 e 2x 2c 1 xe 2x xe 2x e 2x 2xe 2x e 2x 2e 2x c 1 c 2 If r 1 = r 2 = r, the general solution is y(x) = c 1 e rx + c 2 xe rx Example 2.31 16y 4y + 25y = y() = 3, y () = 9 /4 16m 2 4m + 25 = (4m 5) 2 = m = 5 /4 32

y(x) = c 1 e 5x/4 + c 2 xe 5x/4 y (x) = 5 /4 c 1 e 5x/4 + c 2 e 5x/4 + 5 /4 c 2 xe 5x/4 y() = 3 = c 1 + y () = 9 /4 = 5 /4 c 1 + c 2 + c 1 = 3 c 2 = 9 /4 5 3 /4 = 6 y(x) = 3e 5x/4 6xe 5x/4 = 3e 5x/4 (1 2x) Example 2.32 y + 14y + 149y = y() = 1, y () = 8 m 2 + 14m + 149 = m = 7 ± 1i y(x) = c 1 e 7x cos(1x) + c 2 e 7x sin(1x) y (x) = 7c 1 e 7x cos(1x) 1c 1 e 7x sin(1x) 7c 2 e 7x sin(1x) + 1c 2 e 7x cos(1x) = ( 7c 1 + 1c 2 )e 7x cos(1x) + ( 1c 1 7c 2 )e 7x sin(1x) Example 2.33 y + 4y + 13y = y() =, y () = 7 33

m 2 + 4m + 13 = m = 2 ± 3i y(x) = c 1 e 2x cos(3x) + c 2 e 2x sin(3x) y (x) = 2c 1 e 2x cos(3x) 3c 1 e 2x sin(3x) 2c 2 e 2x sin(3x) + 3c 2 e 2x cos(3x) y() = = c 1 y () = 7 = 2c 1 + 3c 2 1 1 1 1 2 3 7 1 3 y(x) = e 2x cos(3x) + 3e 2x sin(3x) MISSING NOTES GO HERE 3 Transformations 3.1 Laplace Transform The Laplace transform of f(t) is denoted by L {f(t)} = f(t)e st dt = F (s) (3.1) 34

Example 3.1 Compute L {e at } F (s) = e at e st dt = e (a s)t dt [ 1 = a s e(a s)t = 1 a s ] [ e e ], s > a = 1 a s = 1 s a, s > a Example 3.2 Compute L {sin(at)} F (s) = sin(at)e st dt u = e st, du = se st IBP v = cos(at), dv = sin(at) dt a F (s) = cos(at)e st s cos(at)e st dt a a u = e st, du = se st dt v = sin(at), dv = cos(at) dt a F (s) = cos(at)e st s ( sin(at)e st + s ) sin(at)e st dt a a a a = cos(at)e st s ( sin(at)e st + s ) a a a a F (s) = cos(at)e st s a a 2 sin(at)e st s2 a F (s) ) 2 F (s) (1 + s2 = (cos(at) e st + s ) a 2 a a sin(at) F (s) = [ (cos(at) e st + s ) ] a a sin(at) a 2 a 2 + s 2 = ( 1a ) a 2 (1 + ) a 2 + s, s > 2 = a a 2 + s 2, s > 35

Properties 1. L{f(t) + g(t)} = L{f(t)} + L{g(t)} 2. L{c f(t)} = c L{f(t)} Example 3.3 Compute L{3 sin(2t)} L{3 sin(2t)} = 3 L{sin(2t)} = 3 2 s 2 + 4, s > = 6 s 2 + 4, s > Example 3.4 Compute L{e t e 2t } L{e t e 2t } = L{e t } L{e 2t } = 1 s 1 1 s 2, s > 2 1 = (s 2)(s 1), s > 2 3.2 Inverse Laplace Transform L{f(t)} = F (s) L 1 {L{f(t)}} = L 1 {F (s)} (3.2) Properties 1. L 1 {F (s) + G(s)} = L 1 {F (s)} + L 1 {G(s)} 2. L 1 {c F (s)} = c L 1 {F (s)} 36

Example 3.5 Compute L 1 { 3 s } } } L 1 { 3 s = 3 L 1 { 1 s = 3(1) = 3 Example 3.6 { } { } 4 L 1 = 4 L 1 1 s + 1 s ( 1) = 4e t Example 3.7 { } { } L 1 2 = 2 L 1 1 s 2 + 3s + 2 (s + 1)(s + 2) A s + 1 + B s + 2 = 2 (s + 1)(s + 2) A(s + 2) + B(s + 1) = 2 Set s = 2 B = 2, A = 2 { } { } f(s) = 2 L 1 1 + ( 2) L 1 1 s ( 1) s ( 2) = 2e t 2e 2t Do for homework: 37

Example 3.8 { } 2s + 4 f(t) = L 1 s 2 + 2s + 5 s 2 + 2s + 5 = s 2 + 2s + 1 + 4 = (s + 1) 2 + 4 = (s + 1) 2 + 2 2 { } 2s + 4 f(t) = L 1 (s + 1) 2 + 2 2 From the table: { } e at sin(bt) = L 1 b, s > a (s a) 2 + b { 2 } e at cos(bt) = L 1 s a, s > a (s a) 2 + b { 2 } 2(s + 1) + 2 f(t) = L 1 (s + 1) 2 + 2 { 2 } = 2 L 1 s + 1 (a = 1, b = 2) (s + 1) 2 + 2 { 2 } + L 1 2 (a = 1, b = 2) (s + 1) 2 + 2 2 = 2e t cos(2t) + e t sin(2t) Theorem 3.9 L {f (t)} = s L {f(t)} f() 38

Proof L {f (t)} = f (t)e st dt u = e st, du = se st dt v = f(t), dv = f (t) dt = [ e st f(t) ] ( s)e st f(t) dt ( = ) e s f( ) e s f() + s = f() + 5 L {f(t)} f(t)e st dt Example 3.1 L {f (t)} = L { [f (t)] } = s L {f (t)} f () = s (s L {f(t)} f()) f () = s 2 L {f(t)} sf() f () Example 3.11 L{f (t)} = L{[f (t)] } = s L{f (t)} f () = s [ s 2 L {f(t)} sf() f () ] f () = s 3 L{f(t)} s 2 f() sf () f () 1. L{f (t)} = s L{f(t)} f() 2. L{f (t)} = s 2 L{f(t)} sf() f () 3. L{f (t)} = s 3 L{f(t)} s 2 f() sf () f () 39

4. L{f (n) (t)} = s n L{f(t)} s n 1 f() s n 2 f ()... f (n 1) () = s n L{f(t) + n i=1 sn i f i 1 () Example 3.12 Solve y = y y() = 1 or dy = y dx y() = 1 1 y dy = dx ln y = (x + c 1 ) y = e x+c 1 = e c 1 e x y = ±e c 1 e x = Ce x y = Ce x but y() = 1 1 = Ce C = 1 y = e x 4

Now let s do it again using the Laplace transform y (x) = y(x) L{y (x)} = L{y(x)} s L{y(x)} y() = L{y(x)} s L{y(x)} L{y(x)} = y() L{y(x)}(s 1) = 1 L{y(x)} = 1 s 1 { } 1 L 1 {L{y(x)}} = L 1, a = 1 s 1 y(x) = e x Example 3.13 Solve y + 2y + 5y = y() = 2, y () = L{y } + 2 L{y } + 5 L{y} = L{} s 2 L{y} sy() y () + 2 (s L{y} y()) 5 L{y} = L{y} ( s 2 + 2s + 5 ) = 2s + 4 L{y} = 2s + 4 s 2 + { 2s + 5 } 2s + 4 L 1 {L{y}} = L 1 s 2 + 2s + 5 y(x) = 2e x cos(2x) + e x sin(2x) Example 3.14 Solve: y + y = sin(2t), y() = 2, y () = 1 41

L {y } L {y} = L {sin(2t)} s 2 L {y} sy() y () + L {y} = 2 s 2 + 4, s > s 2 L {y} 2s 1 + L {y} = 2 s 2 + 4 L {y} (s 2 + 1) 2s 1 = 1 s 2 + 4 L {y} (s 2 + 1) = 1 s 2 + 4 + 2s + 1 = 2 + 2s3 + 8s + s 2 + 4 s 2 + 4 L {y} = 2s3 + s 2 + 8s + 6 (s 2 + 4)(s 2 + 1) { } 2s y = L 1 3 + s 2 + 8s + 6 (s 2 + 4)(s 2 + 1) 2s 3 + s 2 + 8s + 6 (s 2 + 4)(s 2 + 1) = as + b s 2 + 1 + cs + d s 2 + 4 2s 3 + s 2 + 8s + 6 = (as + b)(s 2 + 4) + (cs + d)(s 2 + 1) a + c = 2 b + d = 1 4a + c = 8 = as 3 + bs 2 + 4as + 4b + cs 3 + ds 2 + cs + d = s 3 (a + c) + s 2 (b + d) + s(4a + c) + (4b + d) 4b + d = 6 1 1 2 1 2 1 1 1 1 5/3 4 1 8 1 4 1 6 1 2 /3 2s 3 + s 2 + 8s + 6 = 2s 3 + s 2 ( 5 /3 2 /3) + 8s + 2 /3 2 /3 42

3.3 Unit Step-function Definition The function u(t) is called the unit step-function., t < u(t) = 1, t > (3.3) u c (t) is a transformation of the unit step-function, t c < u c (t) = u(t c) = 1, t c > (3.4) 43

Example 3.15 For a < b, find u(t a) u(t b)., (t < a) (t > b) u(t a) u(t b) = 1, a < t < b Example 3.16 3, t < 2 1, 2 < t < 5 Draw f(t) = t, 5 < t < 8 t 2 /1, t > 8 Theorem 3.17 For C L {u c (t)} = L {u(t c)} = e cs s, s > 44

Proof L {u c (t)} = u c (t)e st dt, t < c Recall u c (t) = 1, t > c c = u c (t)e st dt + u 1 c (t)e st dt c = + e st dt c ] = [ e st s c c = 1 [ ] e sc s = e sc s, s > Example 3.18 L {u c (t)f(t c)} = e cs F (s), s > = u c (t)f(t c)e st dt c = u c (t)f(t c)e st dt + u 1 c (t)f(t c)e st dt c = f(t c)e st dt c Let v = t c = = f(v)e s(v+c) dv f(v)e sv e sc dv = e sc f(v)e sv dv = e sc L {f(v)} = e sc F (s) 45

Example 3.19 For homework: L { t 2 u(t 1) } = L { t 2 u 1 (t) } MISSING NOTES HERE Example 3.2 { t } F (s) = L e (t τ) sin τ dτ = L { e t sin t } = L { e t} L {sin t} ( ) ( ) 1 1 = s + 1 s 2 + 1 1 = (s + 1)(s 2 + 1), s > Example 3.21 { t } F (s) = L sin(t τ) cos τ dτ = L {sin t cos t} ( ) ( ) 1 s = s 2 + 1 s 2 + 1 s = (s 2 + 1), s > 2 46

Example 3.22 { } f(t) = L 1 1 s 2 (s 2 + 1) { } 1 L 1 = t s { 2 } 1 L 1 = sin t s 2 + 1 L {t} = 1 s 2 L {sin t} = 1 ( s 2 + ) 1 ( ) 1 1 L {t} L {sin t} = s 2 s 2 + 1 1 L {t sin t} = s 2 (s 2 + 1) { } t sin t = L 1 1 s 2 (s 2 + 1) t = (t τ) sin τ dτ u = t τ, u = dτ v = cos τ, v = sin τ t = [ cos τ(t τ)] t cos τ dτ = t [sin τ] t = t sin t Example 3.23 { } G(s) L 1 s 2 + 1 F (s) = 1 s 2 + 1 f(t) = sin t { } G(s) sin t g(t) = L 1 s 2 + 1 = t sin(t τ)g(τ) dτ 47

Example 3.24 y + 4y = g(t); y() = 3, y () = 1 L {y } + L {4y} = L {g(t)} s 2 L {y} 3s + 1 + 4 L {y} = G(s) L {y} (s 2 + 4) 3s + 1 = G(s) G(s) + 3s 1 L {y} = s 2 + 4 Let F (s) = 1 s 2 + 4 f(t) = 1 2 sin(2t) L 1 {G(s)F (s)} = 1 sin(2t) g(t) 2 y = 1 { 3s 2 sin(2t) g(t) + L 1 s 2 + 4 1 } s 2 + 4 = 1 2 sin(2t) g(t) + 3 cos(2t) 1 2 sin(2t) = 1 2 t g(t τ) sin(2τ) dτ + 3 cos(2t) 1 2 sin(2t) 3.4 Gamma Function Γ(a + 1) = t a e t dt, a > 1 (3.5) Theorem 3.25 Γ(a + 1) = aγ(a) 48

Example 3.26 Γ(4) = 3Γ(3) = 3 2Γ(2) = 3 2 1Γ(1) = 3 2 1 1 = 6 = 3! So for positive integers, Γ(a + 1) = a! Proof Γ(a + 1) = t a e t dt u = t a, u = a t a 1 v = e t, v = e t [ = t a e t] + a = + aγ(a) t a 1 e t dt 49

Example 3.27 Γ(1) = Γ( + 1) a = = = t e t dt e t dt = [ e t] = 1 Γ(3) = Γ(2 + 1) = 2Γ(2) = 2 1 = 2 Γ(4) = Γ(3 + 1) = 3Γ(2 + 1) = 3 2Γ(1 + 1) = 3! Γ(n + 1) = n!, n Z + 5

Proof [ Γ [ Γ Γ ( 1 2 ( ) 1 2 = Γ ( 12 ) + 1 = t 1 /2 e t dt )] 2 ( ) 2 = t 1 /2 e t dt Let u 2 = t, 2u du = dt ( )] 2 ( 1 = u 1 e u2 2u du 2 = ( 2 ) 2 e u2 du ) 2 ( ) 2 = e u2 du ( ) ( ) = e x2 dx e y2 dy = e (x2 +y 2) dx dy In cylindrical coordinates: x 2 + y 2 = r 2 = 2π e r2 r dr dθ (Let w = r 2, dw = 2r dr) = 1 2 2π 2π = 1 2 [ ( )] 2 1 Γ = 1 2 2 (2π) = π ( ) 1 Γ = π 2 1 dθ e w dw dθ 51

Example 3.28 Γ Γ ( ) ( ) 3 1 = Γ 2 2 + 1 = 1 ( ) 1 2 Γ 2 π = ( ) 7 = Γ 2 2 ( 5 2 + 1 ) = 5 2 Γ ( 3 2 + 1 ) = 5 2 3 2 Γ = 15 4 1 2 Γ = 15 π 8 ( ) 1 2 + 1 ( ) 1 2 Example 3.29 ( Γ 5 ) = Γ ( 83 ) 3 + 1 = 8 ( 11 3 Γ = 88 9 Γ ) 3 + 1 ) ( 14 3 + 1 Notice that it s just getting more and more complicated. So we need to do the opposite of what we ve 52

been doing 5 ( 3 Γ 5 ) = Γ ( 53 ) 3 + 1 ( Γ 5 ) = 35 ( 3 Γ 2 ) 3 2 ( 3 Γ 2 ) = Γ ( 23 ) 3 + 1 = Γ ( Γ 2 ) = 3 ( ) 1 3 2 Γ 3 ( Γ 5 ) = 9 ( ) 1 3 1 Γ 3 ( = Γ 2 ) 3 ( ) 1 3 Example 3.3 ( Γ 13 ) = 713 ( 7 Γ = 49 78 Γ ( 1 7 6 7 ) ) 4 Systems of Differential Equations Example 4.1 dx dt = 3x + 5y dy dt = 3x + y x() = 1, y() = 2 53

(s 3) 5 L {x} = 1 3 (s 1) L {y} 2 L {x (t)} = 3 L {x} + 5 L {y} L {y (t)} = 3 L {x} + L {y} s L {x} 1 x() = 3 L {x} + 5 L {y} s L {y} 2 y() = 3 L {x} + L {y} (s 3) L {x} 5 L {y} = 1 3 L {x} + (s 1) L {y} = 2 5 L {x} (s 3) = L {y} 3 (s 1) 1 1 2 1 (s 1) 5 = s 2 1 4s 12 3 (s 3) 2 = 1 s 2 4s 12 s + 9 L {x} = s 2 4s 12 2s 3 L {y} = s 2 4s 12 x = 1 ( 8 e 2t 15e 8t 7 ) y = 1 8 e 2t ( 9e 8t + 7 ) (s + 9) (2s 3) Example 4.2 x (t) = 3x 5y y (t) = x + 5y x() = 3, y() = 4 54

(s 3) 5 L {x} = 3 1 (s 5) L {y} 4 L {x} = L {y} L {x (t)} = 3 L {x} 5 L {y} L {y (t)} = L {x} + 5 L {y} 3 s L {x} x() = 3 L {x} 5 L {y} s L {y} 4 y() = L {x} + 5 L {y} (s 3) L {x} + 5 L {y} = 3 L {x} + (s 5) L {y} = 4 = 1 s 2 8s + 2 1 s 2 8s + 2 L {x} = 3s 5 s 2 8s + 2 4s 15 L {y} = s 2 8s + 2 (s 5) 5 3 1 (s 3) 4 3s 5 4s 15 x = 1 2 e4t (17 sin(2t) + 6 cos(2t)) y = 1 2 e4t (sin(2t) + 8 cos(2t)) Example 4.3 e x2 dx We can t integrate this using the techniques we learned in Calculus, but we can approximate it as a power series f(x) = n= a n (x a) n, a n = f (n) (a) n! So we can write f(x) as f(x) = a + a 1 (x a) + a 2 (x a) 2 + a 3 (x a) 3 +... + a n (x a) n 55

Example 4.4 y xy =, y() = 1, y () = 2 y(x) = a n x n n= = y() + y () 1! x + y () 2! x 2 + y () x 3 +... 3! Now, we know y() and y (), but nothing higher, so first, let s plug in our initial conditions to the original equation to find y () y (x) xy(x) = y () 1 y() = y () = Now let s take the derivative of the equation, and plug in our initial conditions to find y () y (x) y(x) xy (x) = y () 1 y() 2 y () = y () = 1 Now let s do it for the fourth derivative y (4) (x) y (x) y (x) xy (x) = y (4) (x) 2y (x) xy (x) = y (4) () 2 2 y () 1 y () = y (4) () = 4 56

We can keep going, but we ll stop here with our approximation, giving us y(x) = 1 + 2x + 1 6 x3 + 4 24 x4 +... Example 4.5 y x 2 x =, y() = 1, y () = y (x) = y (x) = y (x) x 2 y = = = n a n x n 1 n=1 n(n 1)a n x n 2 n=2 n(n 1)a n x n 2 x 2 n=2 n(n 1)a n x n 2 n=2 n= a n x n a n x n+2 = n= (n + 2)(n + 1)a n+2 x n n= = (2)(1)a 2 + (3)(2)a 3 x + = 2a 2 + 6a 3 x + a n 2 x n = n=2 (n + 2)(n + 1)a n+2 x n n=2 x n [(n + 2)(n + 1)a n+2 a n 2 ] = n=2 2a 2 =, 6a 3 =, (n + 2)(n + 1)a n+2 a n 2 =, n 2 a n+2 = a n 2 (n + 2)(n + 1), n 2 a n 2 x n = n=2 Example 4.6 Solve ( 4x 2 1)y + 5xy = ; y() = 3, y () = 8 Recall f (n) (a)(x a) n n= n! = a n (x a) n n= We already know y() and y (), so it would be simplest to let a =, as we need to know y(a) and y (a). We re also going to need further derivatives, so lets find y () by evaluating the equation with x =. ( 1)y () + = y () = 57

Now we need the third derivative 8xy + ( 4x 2 1)y + 5y + 5xy = and evaluate at x = + ( 1)y () + 5( 8) + = y () = 4 Now let s use the power series y(x) = y()! + y () 1! x + y () x 2 +... 2! = 3 8x 2 3 x3 +... Example 4.7 Solve ( 3x + 6)y + 8y + (2x 5)y = ; y() = 5, y () = First, evaluate at x = to find y () 6y () + 5 5 y () + ( 5) y() = y () = 25 6 Now take the derivative and evaluate at x = to find y () 3y (x) + ( 3x + 6)y (x) + 8y (x) + 2y(x) + (2x 5)y (x) = 3 25 /6 y () + 6y () + 8 25 /6 y () + 2 y(x) 5 5 y (x) = 75 6 + 6y () + 1 3 + 1 = y () = 185 36 58

We can go further, but if we stop here, we have y(x) = y() + y () 1! x + y () 2! x 2 + y () x 3 +... 3! y(x) = 5 + 25 12 x2 185 216 x3 +... Example 4.8 Solve xy + y + y = ; y() = 1, y () = 1 Radius of Convergence Definition Let p(x), q(x), and r(x) be analytic functions. A point x is a singular point for the differential equation p(x)y (x) + q(x)y (x) + r(x)y(x) = (4.1) if q(x ) p(x ) or r(x ) p(x ) is undefined. Otherwise, x is called a regular point. Try solving the following: xy + y + y = y() = 1, y () = 1 If we try evaluating at x = to find y (), we end up multiplying the y () by zero, so we can t get it by that method. We could rewrite this problem by dividing everyting by x y + 1 x y + 1 x y = Now we see that x = is a singular point. Find the singular points for the following equations: x 2 y + xy = x = 2y + y x 1 = x = 1 59

Theorem 4.9 Suppose y(x) = a n (x x ) n n= is a series solution do Equation 4.1. Then, the radius of convergence for y(x) is at least as large as the distance from x (regular point) to the nearest singular point of the equation. Example 4.1 Find the radius of convergence about x (x + 1)y + y + (x + 1)y =, about x = So first we divide by (x + 1) y + y x + 1 + y = x = 1 is a singular point. So the radius of convergence is at least 1. Example 4.11 Find the radius of convergence about x (x 2 2x 3)y + (x + 1)y + 3y =, about x = First divide by (x+1), to get singular point x = 1. Then divide the original equation by (x 2 2x 3) to get y + (x + 1)y (x 3) (x + 1) + 3y (x 3)(x + 1) = which has singular points x = 1 and x = 3. So the radius of convergence is at least 1. Example 4.12 (x 3 + x 2 + x + 1)y + 2xy + (x 1)y =, about x = Factor the big polynomial first x 3 + x 2 + x + 1 = x 2 (x + 1) + 1(x + 1) = (x 2 + 1)(x + 1) 6

Now divide that out to get y + 2xy (x 2 + 1)(x + 1) + (x 1)y (x 2 + 1)(x + 1) = x = 1, x = ± ι. Radius of convergence is at least 2 61

5 Final Examples Example 5.1 L {t cosh(t)} cosh(t) = et + e t 2 F (s) = 1 2 L { te t + te t} = 1 2 L { te t} + 1 2 L { te t} L { t n e at}, n Z + n! = (s a), s > a n+1 F (s) = 1 1! 2 (s 1) + 1 1! 1+1 2 (s + 1), s > 1 1+1 = 1 ( ) 1 2 (s 1) + 1, s > 1 2 (s + 1) 2 Example 5.2 L {f(t)}, where f(t) = 3 cos(2t) + t e 2(t τ) τ sin(3τ) dτ { t } L {f(t)} = 3 L {cos(2t)} + L e 2(t τ) τ sin(3τ) dτ s = 3 s 2 + 4 + L { e 2t t sin(3t) } s = 3 s 2 + 4 + L { e 2t} L {t sin(3t)} s = 3 s 2 + 4 + 1 6s s 2 (s 2 + 9) 2 s = 3 s 2 + 4 + 6s (s 2)(s 2 + 9) 2 62

Example 5.3 { } 2s + 9 L 1 s 2 + 8s + 17 { } { } f(t) = L 1 2s + 9 2s + 9 = L 1 s 2 + 8s + 16 + 1 (s + 4) 2 + 1 { } { 2 } = 2 L 1 s + 9 L 1 1 (s + 4) 2 + 1 2 (s + 4) 2 + 1 { } { 2 } s + 4 4 = 2 L 1 + 9 L 1 1 (s + 4) 2 + 1 2 (s + 4) 2 + 1 { } { } 2 = 2 L 1 s + 4 + L 1 1 (s + 4) 2 + 1 2 (s + 4) 2 + 1 2 = 2e 4t cos(t) + e 4t sin(t) = e 4t (2 cos(t) + sin(t)) Example 5.4 y + 6y + 8y = δ(t), y() =, y () = s 2 L {y} + 6s L {y} + 8 L {y} = e s L {y} (s 2 + 6s + 8) = 1 1 L {y} = s 2 + 6s + 8 1 = (s + 2)(s + 4) 1 (s + 2)(s + 4) = A s + 2 + B s + 4 1 = A(s + 4) + B(s + 2) A = 1 /2, B = 1 /2 L {y} = 1 1 2 s + 2 1 1 2 s + 4 y = 1 ( e 2t e 4t) 2 63

Example 5.5 y + 2xy + 4y =, y() = 1, y () = 2 The solution is: Just use: n= y (n) () x n n! y(x) = 1 + 2x 2x 2 2x 3 + 4 3 x4 +... Example 5.6 Find the radius of convergence for the series solution about x = to the equation: (x 2 + 2x + 1)(x 2 + 6x + 8)y + 2xy 4y = y + 2xy (x + 2)(x + 4)(x 2 + 2x + 1) 4y (x + 2)(x + 4)(x 2 + 2x + 1) = You can actually just use the quadratic formula. The roots will be the singular points, so the singular points are x = 1 ± 3i and x = 4, 2. The closest point is 1 3i, and it is 2 away from, so the radius of convergence is at least 2. Example 5.7 y + 3y + 2y = 1 1 + e x = g λ 2 + 3λ + 2 = λ = 2, λ = 1 y h = c 1 e 2x + c 2 e x = y 1 + y 2 y 1 y 2 W = y 1 y 2 y2 g y p = y 1 W + y y1 g 2 W 64