MATH529 Lesson00: ODE Review

Chapter 1 Introduction to ODE review

ODE:
  Explicit form (y^((n)))=f(x,y,y',..,(y^((n-1))))
  Implicit form  0=F(x,y,y',..,(y^((n-1))),(y^((n))))
 First order ODEs, n=1, explicit form y'=f(x,y)
 Separable first-order ODE y'=g(x)h(y)=((d y)/(d x)), g(x)d x=((d y)/(h(y)))

Separable ODEs

p.30, Ex. 7 x y'=2 y, y'=2 (y/x), g(x)=(1/x), h(y)=2y, (1/x)d x=((d y)/(2 y))
Solve by integration on both side ∫(1/x)dx=∫(1/(2y))dy, log(x)=(1/2) log(y)-C

g[x_]=(1/x); h[y_]=2y;

Integrate[g[x],x]

Integrate[1/h[y],y]

DiffEq = D[y[x],x] == g[x] h[y[x]]

DSolve[ DiffEq,y[x],x]

p.30, Ex 11, y''+9y=18, y''=18-9y
Homogeneous linear ODE of order n, (y^((n)))+(Subscript[a, n-1])(y^((n-1)))+(Subscript[a, n-2])(y^((n-2)))+...+(Subscript[a, 1])y'+ (Subscript[a, 0]) y=0 
Guess solution to be of form y(x)=(e^(r x)), obtain ((r^n)+(Subscript[a, n-1])(r^(n-1))+(Subscript[a, n-2])(r^(n-2))+...+(Subscript[a, 1])r+ (Subscript[a, 0]))(e^(r x))=0, implying that (Subscript[p, n])(r)=(r^n)+(Subscript[a, n-1])(r^(n-1))+(Subscript[a, n-2])(r^(n-2))+...+(Subscript[a, 1])r+ (Subscript[a, 0])=0

Step 1: Solve the homogeneous problem y''+9y=0, characteristic equation, (r^2)+9=0, with roots (Subscript[r, 1])=3i, (Subscript[r, 2])=-3i. Solution is (Subscript[y, h])(x)=(Subscript[c, 1]) (e^(3i x))+(Subscript[c, 2]) (e^(-3i x))=A cos(3x)+B sin(3x)

Step 2: State that the solution is the sum of the homogeneous solution and a particular solution y=(Subscript[y, h])+(Subscript[y, p])

DiffEqH = D[y[x],{x,2}] + 9 y[x] == 0

DSolve[DiffEqH,y[x],x]

DiffEq = D[y[x],{x,2}] + 9 y[x] == 18

DSolve[DiffEq,y[x],x]

p.30, Ex. 12.  x y''-y'=0, Let y'=z, x z'=z, a separable ODE  x (dz/dx)=z, x (dz/z)=(dx/x) with solution z(x)=C x, and y(x)=(1/2)C (x^2)+D

DiffEq = x y''[x] - y'[x] == 0

DSolve[ DiffEq,y[x],x]

p.31, Ex. 34, (1+x y)y'+(y^2)=0. Consider y=(e^(-x y))

Subst = y[x] == Exp[-x y[x]]

result = Simplify[D[Subst,x]]

verif = result /. (E^(-x y[x]))-> y[x]

DiffEq = (1+ x y[x]) y'[x] + (((y[x]))^2) == 0

DSolve[DiffEq,y[x],x]

?ProductLog