ODEs

Question 1

Linear

Answer 1

Does not contain the products of any functions of the unknown function or its derivatives

Question 2

Homogeneous

Answer 2

All terms contain the unknown function

Question 3

Methods for solving 1st order homogeneous ODEs

Answer 3

• Separable equation
• Separation of variables
• Introduce new variable = y(x)/x

Question 4

Methods for solving 1st order non-homogeneous equations

Answer 4

• Integrating factor
• Variation of parameter

Question 5

Methods for solving 2nd order homogeneous ODEs

Answer 5

• Constant coefficients method

Question 6

Methods for solving non-homogeneous equations

Answer 6

• Reduction of order
• Variation of parameter

Question 7

Separable equation method

Answer 7

General form: y’ = g(x)h(y)
- Divide by h(y) make sure to work out the solution this division gives you.
- Integrate both sides as dy/dx and dx cancel to give you dy/h(y).
- Manipulate to find y.

Question 8

Equations reducible to separable form method

Answer 8

General equation: y’ = f(y/x)
- Introduce v(x) = y(x)/x
- y(x) = xv(x)
- Product rule to get y’ = v + xv’
- Sub new form for y’ and v bacj in to original equation
- Manipulate to find v’
- This equation has now been reduced to separable form

Question 9

1st order variation of parameter method

Answer 9

General equation: y’ + a(x)y = f(x)
- Solve the corresponding homogeneous equation by separable equation method: y0’ + a(x)y0 = 0
- Use A’(x) = -a(x) to get y0 = ce^(A(x)) where c = +/- e^c
- Let y(x) = C(x)e^(A(x))
- Use product rule to find y’, subbing a for A’
- Sub all terms back into original equation and cancel to find C’ = e^(-A)f
- Integrate for result

Question 10

General solution for constant coefficients method with two distinct roots

Answer 10

y = Ae^(ų1x) + Be^(ų2x)

Question 11

General solution for constant coefficients method with repeated root

Answer 11

y = (A + Bx)e^(ųx)

Question 12

General solution for constant coefficients method for complex roots

Answer 12

ų1, ų2 = a +/- ig
y = e^(ax)(Acos(gx) + Bsin(gx))

Question 13

Solving Euler’s Equation

Answer 13

General equation: x^(2)y” + axy’ + by = 0
- let y = x^(ų) where ų is a constant.
- find y’ and y” and sub into original equation.
- x^(ų) cancels leaving a quadratic to be solved using constant coefficients method

Question 14

Reduction of order method

Answer 14

General equation: y” + p(x)y’ + q(x)y = f(x)
- Suppose that a solution of the equation is known, y1(x):
y1” + p(x)y1’ + q(x)y1 = 0
- The solution now has the form y(x) = u(x)y1(x) where u(x) is unknown therefore reducing the equation to a 1st order ODE

Question 15

2nd order variation of parameter method

Answer 15

• write equation in y0 form that equals 0
• sub in ų and find roots to get y0 = c1e^(ų1x) + c2e^(ų2x)
• let y equal this and find y’ this will result in one equation for c1, c2 but to determine them uniquely we are free to choose another relation.
• let c1’e^(ų1x) + c2’e^(ų2x) = 0
• Find y” from now cancelled down y’
• sub y” and y’ and y into original equation, cancel and manipulate to find c1 and c2
• solution is: y = c1(x)y1(x) + c2(x)y2(x)

Question 16

Method of undetermined coefficients

Answer 16

General equation: y” + ay’ + by = f(x)
- y = yc(x) + yp(x)
- yc” + ayc’ + byc = 0, yp(x) = ?

1) if f(x) is a polynomial, try yp = polynomial of the same order

2) if f(x) = sin(kx) or cos(kx), try yp = asin(kx) + Bcos(kx)

3) if f(x) = xe^(kx), try yp = Axe^(kx)