Sem 2 Stewart Flashcards
method for 1st order differential equation
separating the variables
integrating factor if variables cannot be seperated
method for second order homogeneous differential equations
auxiliary equation and then general solution depending on roots of auxiliary equation
method for second order non-homogeneous differential equations
complementary function and particular integral in the same form as f(x) or multiply by x or x^2 if in same form as CF
form of general solution for real unequal roots
y=Ae^ax+Be^bx
form of general solution for real equal roots
y=(Ax+B)e^ax
form of general solution for complex roots
m=a+/-bi
y=e^ax(Acosbx+Bsinax)
steps for Riemann sums
- divide domain into n strips of width Δx=b-a/n
- find grid points
- evaluate riemann sum
(could be left, right, upper or lower)
midpoint rule
Δx[f(x1bar)+…+f(xnbar)]
where Δx=(b-a)/n and x bar is midpoint of interval
trapezium rule
the average of the L and R Riemann sums
1/2(Ln+Rn)
=Δx/2[f(x0)+2f(x1)+…+2f(xn-1)+f(xn)]
simpson’s rule
use parabolae instead of straight lines
n must be even
consider consecutive pairs of intervals and approximate the integral over each pair
Δx/3(y0+4y1+2y2+4y3+2y4+…+2yn-2+4yn-1+yn)
general steps for approximate integration
- find Δx
- find x0,x1,…,xn
- (midpoint) x1bar, x2bar…
- f(x1), f(x2),… etc
- apply formula
function is continuous on an interval if…
continuous at every number in interval
function is differentiable at a if..
f’(a) exists
differentiable on an open interval if differentiable at every point
F is an anti derivative of f if
F’(x)=f(x) for all x
what form are integrals in for integration by substitution?
∫f(g(x))g’(x)
u=g(x) is a differentiable function, continuous on I
∫f(g(x))g’(x) dx = ∫f(u) du
limits in integration by substitution
change limits using expression for u and sub in current limits
limits are g(a) and g(b)
trigonometric integrals
manipulate integrand using trig identities
can use substitution in reverse by setting x=g(f) and obtain inverse substitution
integral on closed interval [-a,a], if f is even (i.e. f(-x)=f(x))
∫a/-a f(x)dx = 2∫a/0 f(x) dx
integral on closed interval [-a,a], if f is odd (i.e. f(-x)=-f(x))
∫a/-a f(x)dx = 0
integration by parts formula
∫uv’=uv-∫vu’
what is needed to split into partial fractions?
proper function
degree numerator < degree denominator
if not, denominator into numerator