Updated 2013

1.2-2  y'=(x-2)^2, y(2)=1
=====

 Details: See Jennifer Lahti's solution, reference below.

 The METHOD OF QUADRATURE is not referenced in the textbook. See
 however Maple's documentation on methods of solutions for ordinary
 differential equations, where the first and primary method is the
 method of quadrature. The textbook references in maple are a record
 not only of the origins of the methods, but the naming conventions
 as well.

 QUADRATURE means INTEGRATE and that is the method, using the
 Fundamental Theorem of Calculus to evaluate integrals. The plan is
 to integrate both sides of the equation and determine a candidate
 solution to the differential equation.

 The textbook famously omits most steps in the examples, making it
 difficult to identify the method and the theory applied. See
 Example 1, section 1.2, for such an example.

 FUNDAMENTAL THEOREM OF CALCULUS
 Assume f is continuously differentiable and g is continuous.
 Then

 DEFINITE INTEGRALS
 (a) int(f'(x),x=a..b) = f(b)-f(a)
 (b) (d/dx)int(g(t),t=a..x)=g(x)
 INDEFINITE INTEGRALS
 (a) int(f') = f(x)+constant
 (b) (d/dx)int(g)=g(x)

 Reference: Fund Thm Calculus and method of quadrature examples
 http://www.math.utah.edu/~gustafso/FTC-Method-of-Quadrature.pdf
 
 Reference: Jennifer Lahti's solution to 1.2-2
http://www.math.utah.edu/~gustafso/BackgroundLogExp-decay-law-derivation.pdf

1.2-4  dy/dx = 1/x^2, y(1)=5
=====

 Start with the method of quadrature to get int(dy/dx) = int(1/x^2),
 both indefinite integrals. Use the Fundamental theorem of calculus
 on the left and the power rule for integrals on the right.

 An answer check is expected. Either cite the back of the book or do
 a complete answer check, handwritten. See Jennifer Lahti's answer
 check to 1.2-2, above, for a sample of the expected details.

1.2-6  dy/dx = x sqrt(x^2+9), y(-4)=0
=====

 Start with the method of quadrature to get 
    int(dy/dx) = int(x sqrt(x^2+9)),
 both indefinite integrals. Use the Fundamental theorem of calculus
 on the left and u-substitution for integrals on the right.
 
 Choices for a u-substitution are not many; settle on u=x^2+9 and
 du=2x dx. Then
    int(dy/dx) = int(x sqrt(x^2+9))
               = int((1/2) sqrt(u))
 The last integral is evaluated with the power rule for integrals,
 because sqrt(u) = u^n with n=1/2.
               
 An answer check is expected. Either cite the back of the book or do
 a complete answer check, handwritten. See Jennifer Lahti's answer
 check to 1.2-2, above, for a sample of the expected details.



1.2-10  dy/dx = x exp(-x), y(0)=1
======

 Start with the method of quadrature to get 
    int(dy/dx) = int(x exp(-x)),
 both indefinite integrals. Use the Fundamental theorem of calculus
 on the left and u-substitution for integrals on the right.
 
    int(dy/dx) = int(x exp(-x)),
               = int(-u exp(u)),    using u=-x, du=-dx 
               = -int(u exp(u)),    a lookup in the integral table, 
                                    on the back inside book cover
               = -((u-1)exp(u)+c,   Table entry 46
               = -((-x-x)exp(-x)+c, use u=-x
               = (x+1)exp(-x)+c                  

 Evaluate integration constant c using y(0)=1. Then check the
 answer. See also online solutions to 1.2-10, with more details.