reduction of order & 3 cases method by khairul anwar

KNF1023 Engineering Mathematics II

Week 8 Second-Order Differential Equations

Chapter Contents:  Preliminary Theory on Linear Equations  Reduction of Order  Homogeneous Linear Equations with Constant Coefficients  Undetermined Coefficients  Variation of Parameters  Cauchy-Euler Equation

Week 7 Preliminary Theory on Linear Equations Theory of Solutions of y” + p(x)y’ + q(x)y = f(x) From

y ' ' p( x) y ' q ( x) y  f ( x)

We can define the initial value problem to be the differential equation, defined on some interval, together with TWO initial conditions: 1. one specifying a point lying on the solution curve 2. one specifying its slope at that point. This problem has the form

y ' ' p( x) y ' q( x) y  f ( x); y ( x0 )  A, y ' ( x0 )  B in which A and B are given real numbers.

Week 7 Preliminary Theory on Linear Equations Homogeneous Linear 2nd Order ODEs When f(x) is identically zero in

y ' ' p( x) y ' q ( x) y  f ( x)

the resulting equation y ' ' p ( x) y ' q ( x) y  0 is called homogenous. Homogeneous simply means the right side of the equation is ZERO. Example: 1. 2 y ' '3 y '5 y  0 2. xy' ' y ' xy  0 3. y ' '25 y  e  x cos x

Homogeneous linear 2nd order DE Homogeneous linear 2nd order DE Nonhomogeneous linear 2nd order DE

Week 7 Preliminary Theory on Linear Equations Homogeneous Linear 2nd Order ODEs: Superposition Principle Theorem on Superposition Principle – Homogeneous Equations Let y1 and y2 be solutions of the homogeneous linear second order differential equation on an interval I

y ' ' p( x) y ' q( x) y  0 Then the linear combination y  c1 y1  c2 y2 is also a solution on the interval. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

Note: This theorem does not apply for Nonhomogeneous equations.

Week 7 Preliminary Theory on Linear Equations Homogeneous Linear 2nd Order ODEs To determine whether TWO solutions are linearly independent on an interval, use Wronskian method: For linearly independent solutions

W ( x) ď&#x20AC;˝

y1 ( x)

y2 ( x)

y1 ' ( x)

y2 ' ( x)

for every x in the interval.

ď&#x201A;š0

Week 7 Preliminary Theory on Linear Equations Homogeneous Linear 2nd Order ODEs This gives further definition of solutions: 1. If y1 and y2 are linearly independent, these are called Fundamental Set of Solutions. 2. If y1 and y2 are Fundamental Set of Solutions, then the general solution of the equation is

y ď&#x20AC;˝ c1 y1 ď&#x20AC;Ť c2 y2

Week 7 Preliminary Theory on Linear Equations Nonhomogeneous Linear 2nd Order ODEs

For equation

y ' ' p( x) y ' q ( x) y  f ( x)

the general solution is

  c1 y1  c2 y2  y p

Week 7 Preliminary Theory on Linear Equations Nonhomogeneous Linear 2nd Order ODEs To prove that a function is a general solution for a nonhomogeneous equation: 1. Prove that y1 and y2 are both solutions for the associated homogeneous equation. 2. If both y1 and y2 are the solutions, prove that both form a fundamental set of solutions by checking if they are linearly independent (using Wronskian). 3. If both y1 and y2 are the fundamental set of solutions, then c1y1 + c2 y2 form the general solution for the homogeneous part. 4. Prove that yp is a particular solution for the nonhomogeneous equation.

Week 7 Preliminary Theory on Linear Equations Nonhomogeneous Linear 2nd Order ODEs

In summary, we can solve nonhomogeneous equation y ' ' p( x) y ' q ( x) y  f ( x) by the following strategy: 1. Find the general solution c1 y1  c2 y2 of the associated homogeneous equation y ' ' p( x) y ' q( x) y  0 2. Find any solution yp of y ' ' p( x) y ' q ( x) y  f ( x) 3. Write the general solution c1 y1  c2 y2  y p . This expression contains all possible solutions of the nonhomogeneous differential equation.

Week 8 Second-Order Differential Equations

Week 8 2nd-Order DE: Reduction of Order

Reduction of Order: Given, a homogeneous second-order differential equation y ' ' p( x) y ' q( x) y  0 ___________ (1)

We want to find TWO independent solutions. Reduction of order is a technique for finding a second solution, if the first solution is known.

Week 8 2nd-Order DE: Reduction of Order

Reduction of Order: y ' ' p( x) y ' q( x) y  0 _________ (1)

The basic idea is that equation (1) above can be reduced to a linear first-order DE by means of a substitution involving the known solution y1. A second solution y2 can be found after the firstorder DE is solved.

Week 8 2nd-Order DE: Reduction of Order

Reduction of Order: y ' ' p( x) y ' q( x) y  0 _________ (1)

Suppose we know y1 is a solution of equation (1). Now we want to find y2. For y1 and y2 to form a set of fundamental solutions, y1 and y2 must be linearly independent, i.e. y2 ( x)  constant y1 ( x)

Week 8 2nd-Order DE: Reduction of Order

Reduction of Order: So, let Or,

y2 ( x)  u ( x) y1 ( x ) y2 ( x)  y1 ( x).u( x)

The function u(x) can be found by substituting y2 ( x)  y1 ( x).u( x)

into the given differential equation.

Week 8 2nd-Order DE: Reduction of Order Example 1: Given that y1  e x is a solution of y ' ' y  0 . Use reduction of order to find a second solution y2. Solution: By reduction of order, let

y2  u.y1 y2  uex

Differentiate y2 twice (by Product Rule):

y2 '  u ' e x  ue x y2 ' '  uex  2e xu 'e xu ' ' For y2 to be a solution of the differential equation y ' ' y  0: ( y2 )' 'y2  0

Week 8 2nd-Order DE: Reduction of Order Solution:

( y2 )' 'y2  0

From

ue x  2e x u 'e x u ' 'ue x  0

Substitute

y2 ' '

e x (2u 'u ' ' )  0

Rewrite,

Since y1  e is a solution, e  0 . x

Thus, we need 2u'u' '  0 .

Week 8 2nd-Order DE: Reduction of Order Solution: To solve 2u'u' '  0 , we use substitution.

Let

w  u' :

2u'u' '  0 2w  w'  0 w'2w  0

Which reduced to a linear first-order equation. 2x

By linear equation method of solving, we found the integrating factor, e : And Integrate





d 2x e w 0 dx w  c1e 2 x

Week 8 2nd-Order DE: Reduction of Order Solution: From Substitute back,

w  c1e22xx u '  c1e

Integrates again,

u   c1e2 x dx

1 2 x u   c1e  c2 2

From y2  u.y1:  1 2 x  y2    c1e  c2 e x  2  1 x y2   c1e  c2 e x 2

Week 8 2nd-Order DE: Reduction of Order Solution:

1 x y2   c1e  c2 e x 2 x By choosing c2 = 0 and c1 = -2, y2  e . From

Check that y1 and y2 are linearly independent using Wronskian: x x e e W (e x , e  x )  x  2  0 x e e x y  e Thus, shown that y1  e and 2 are linearly independent x

solutions of a linear second-order equation, thus form a general solution

y  c1e  c2e x

x

Week 8 2nd-Order DE: Reduction of Order

Reduction of Order (by Formula): Second solution y2 for y ' ' p( x) y ' q( x) y  0 can also be obtained using the formula,

e  y2  y1  2 y1

 P ( x ) dx

Week 8 2nd-Order DE: Reduction of Order Example 2 (Finding second solution by formula): Given that y1  x 2 is a solution of x 2 y ''  3xy . '4 y  0 . Find the general solution of the differential equation. Solution: From Rewrite to standard form

x 2 y ''  3xy'4 y  0 3 4 '' y  y ' 2 y  0 x x

e  dx ------------------------- (1) Using the formula y2  y1  2 y1 1 3  dx   P ( x ) dx 2 3 ln x ln x 3 x e e e  x3 where y1  x and e  P ( x ) dx

x3 Substitute into formula (1): y2  x  4 dx x 2 1 y2  x  dx x y2  x 2 ln x (Ignore the constant after integration) 2

Thus, the general solution y  c1 y1  c2 y2 is

y  c1 x  c2 x ln x 2

Week 8 2nd-Order DE: Reduction of Order Sample Final Exam Question (Semester 2 2009/2010): Function y1  e x / 3 is a solution of 6 y ''  y ' y  0 . Find the second solution y2. Solution:

Week 8 2nd-Order DE: Reduction of Order Sample Final Exam Question (Semester 2 2009/2010): Function y1  e x / 3 is a solution of 6 y ''  y ' y  0 . Find the second solution y2. Solution (continue):

Week 8 Second-Order Differential Equations

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients: Recall back our first-order differential equation: If you have a homogeneous linear equation of the first order where coefficients ay'by  0 and both a and b are constants; a  0 You know that you can solve it using separable variable method or linear equation method. BUT, do you know that you can also solve the above DE using simple algebra.

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients: Let us try this,

ay'by  0

b y'   y a b Let, k   : a

y '  ky where k is a constant

Note that the only basic function whose derivative is a mx constant multiple of itself is an exponential function e .

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients: So, the new solution method: and its derivative y '  me ay'by  0 amemx  bemx  0 e mx (am  b)  0

Substitute y  e This gives mx

into

Since e is never zero for real values of x, the above equation can only be satisfied if am  b  0 . mx y  e By solving the value of m, we have obtained which is a solution of the first order differential equation.

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients: Let’s apply this method of solution to a second-order differential equation ____________ (1) ay' 'by' cy  0 where a, b and c are constants. If we try to find a solution of the form y  e , substitute y  e and its derivatives y '  me mx and y ' '  m 2 e mx into (1) gives mx

am2 e mx  bmemx  ce mx  0 mx 2 e (am  bm  c)  0

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients: Again, as e

is never zero for all real value of x

am2  bm  c  0

______________ (2)

Equation (2) above is called the auxiliary equation. When m is chosen as a root of the quadratic equation (2), the two roots will be  b  b 2  4ac m

Or,

 b  b 2  4ac m1  2a

 b  b 2  4ac and m2  2a

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients: Thus, based on the two roots, there will be 3 forms of the general solution of equation ay' 'by' cy  0 corresponding to the 3 cases: Case I: m1 and m2 are real and distinct (b  4ac  0), 2

Case II: m1 and m2 are real and equal (b 2  4ac  0), Case III: m1 and m2 are conjugate complex number

(b  4ac  0) 2

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients CASE I: DISTINCT REAL ROOTS______________ (b 2  4ac  0) Distinct means different or unequal. Under the assumption that m1 and m2 are unequal, we can find the 2 solutions, m1 x m2 x and y2  e y1  e

Verified that y1 and y2 are linearly independent , thus form a fundamental set of solutions. Thus, the general solution is

y  c1e

m1 x

 c2e

m2 x

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients CASE II: REPEATED REAL ROOTS______________ (b 2  4ac  0) When m1 = m2 , we obtain only ONE solution y1

e

m1 x

We can find the second solution from reduction of order   P ( x ) dx formula: e y2  y1  dx 2 y1 From Rewrite

ay' 'by' cy  0 b c y ' ' y ' y  0 a a

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients CASE II: REPEATED REAL ROOTS______________ (b 2  4ac  0) b c y ' ' y ' y  0 a a Find e 

 P ( x ) dx

e





b dx a

 b  b 2  4ac From quadratic equation, m1  2a

b when b  4ac  0 , m1   2a 2

Thus, integrating factor, b   dx 2 m1dx  a e e  e 2 m1 x

b   2m1 a

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients CASE II: REPEATED REAL ROOTS______________ (b 2  4ac  0) Using the formula:

e  y2  y1  dx 2 y1 2 m1 x e y2  e m1x  2 m1x dx e y2  e m1x  dx y2  xem1x

Thus, the general solution is

 P ( x ) dx

y  c1e

m1 x

 c2 xe

m1 x

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients CASE III: COMPLEX ROOTS______________ (b 2  4ac  0) If m1 and m2 are complex, then we can write m1    i and m2    i where α and β are real and i   1 .

From We obtain

Or,

y  c1e

m1 x

y  c1e

(  i ) x

x

 c2e

m2 x

 c2e

( i ) x

x

y  c1e cos x  c2e sin x

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Examples: Solve the following differential equations. (a) 2 y ' '5 y '3 y  0 (b) y ' '10 y '25 y  0

Solutions: (a) 2 y ' '5 y '3 y  0 Let, a = 2, b = -5 and c = -3 and auxiliary equations is 2m2 - 5m - 3 = 0.

b 2  4ac  (5) 2  4(2)(3)  49  0

This is Case I where m1 and m2 are distinct real roots (b  4ac  0). Thus, 2 2  b  b  4ac  b  b  4ac m  2 m1  2a 2a 1 m2   m1  3 2 2

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Examples: Solve the following differential equations. (a) 2 y ' '5 y '3 y  0 (b) y ' '10 y '25 y  0

Solutions: (a) 2 y ' '5 y '3 y  0 For Case I, general solution is

y  c1e

Thus,

y  c1e 3 x  c2 e

m1 x

 c2e

m2 x

1  x 2

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Examples: Solve the following differential equations. (a) 2 y ' '5 y '3 y  0 (b) y ' '10 y '25 y  0

Solutions: (b) y ' '10 y '25 y  0 Let, a = 1, b = -10 and c = 25 and auxiliary equations is m2 - 10m + 25 = 0.

b 2  4ac  (10) 2  4(1)(25)  0

2 This is Case II where m1 and m2 are repeated real roots (b  4ac  0) . Thus, m1 = m2.

m 2  10m  25  (m  5)(m  5)  0 Thus, m1 = m2 = 5

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Examples: Solve the following differential equations. (a) 2 y ' '5 y '3 y  0 (b) y ' '10 y '25 y  0

Solutions: (b) y ' '10 y '25 y  0 For Case II, general solution is

y  c1e

Thus,

y  c1e  c2 xe

m1 x

 c2 xe

m1 x

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Examples: Solve the following differential equations. (a) 2 y ' '5 y '3 y  0 (b) y ' '10 y '25 y  0

Solutions: (c) y ' '4 y '7 y  0 Let, a = 1, b = 4 and c = 7 and auxiliary equations is m2 + 4m + 7 = 0.

b 2  4ac  (4) 2  4(1)(7)  12  0

2 This is Case III where m1 and m2 are complex roots (b  4ac  0)

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Examples: Solve the following differential equations. (a) 2 y ' '5 y '3 y  0 (b) y ' '10 y '25 y  0

Solutions: (c) y ' '4 y '7 y  0 From m 2  4m  7  0 :

 b  b 2  4ac m1  2a  4   12 m1  2  4  2 3i m1  2

 b  b 2  4ac m2  2a  4   12 m2  2  4  2 3i m2  2

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Examples: Solve the following differential equations. (a) 2 y ' '5 y '3 y  0 (b) y ' '10 y '25 y  0

Solutions: (c) y ' '4 y '7 y  0 m2  2  3i

m1  2  3i

which gives

  2

and

 3

For Case III, general solution is Thus,

y  c1e

2 x

y  c1e

(  i ) x

( i ) x

 c2e x x y  c1e cos x  c2e sin x

cos 3x  c2e

2 x

sin 3x

Week 8 2nd-Order DE: Homogeneous Linear Equations with Constant Coefficients Homogeneous Linear Equations with Constant Coefficients: General Solution Case I

m1 and m2 are real 2 b - 4ac > 0 m1 ≠ m2

y  c1e m1x  c2e m2 x

Case II

m1 and m2 are real 2 b - 4ac = 0 m1 = m2

y  c1e m1x  c2 xem1x

m1 and m2 are Case III conjugate complex b2 - 4ac < 0 number

y  c1e( i ) x  c2e( i ) x y  c1ex cos x  c2ex sin x