Exam 1 Flashcards
derivative of tan(x)
= sec^2(x)
derivative of sec(x)
= sec(x)tan(x)
derivative of csc(x)
= -csc(x)cot(x)
derivative of cot(x)
= -csc^2(x)
integration of sec^2(x)
= tan(x)
integration of csc^2(x)
= -cot(x)
integration of sec(x)tan(x)
= sec(x)
integration of csc(x)cot(x)
= -csc(x)
integration of tan(x)
= ln(sec(x))
integration of cot(x)
= -ln(csc(x)) or =ln(sin(x))
Linear functions
functions of independent variable only. the dependent variable and all of its derivatives are of the first degree
no trig, no e^(independent variable), no ln(independent variable), no powers other than 1 on independent variable
Theorem: Existence and Uniqueness of Solution (Given the IVP)
if f and partial y of f are defined for (x(not), y(not)) then the IVP has a unique solution
How to solve linear DE
Linear Problem in standard form is
dy/dx + p(x) y = Q(x)
1) Find integrating factor mu(x) = e^integral(p(dx))
2) (d/dx)[mu(x)y] = mu(x)Q(x)
3) Integrate both sides with respect to x (DONT FORGOT CONSTANT OF INTEGRATION)
4) Rearrange to solve for desired variable
Separable functions
A DE is separable if it can be written as
dy/dx = g(x)p(y)
How to solve separable DE
standard form: dy/dx = g(x)p(y)
- Check if zeros of p(y) are solutions of equation
- divide both sides by p(y) and multiply both sides by dx to get form: (1/p(y))dy = g(x)dx
- Integrate both sides to get an implicit solution
Exact Equations
M(x,y)dx+N(x,y)dy = 0
(partial f or x)dx + (partial f of y)dy = 0
Solving Exact equations
M(x,y)dx + N(x,y)dy = 0
1. Check for exactness partial M of y = partial N of x
2. If it is not exact, solve for integrating factor using
either mu(x) = e^integral ((M partial y - N partial x)/N) dx
or mu (y) = e^integral ((N partial x - M partial y)/M)dy
3. If/when exact f(x,y) = integral M(x,y)dx + g(y)
4. Take partial with respect to y
5. Set g’(y) = N and integrate
6. Solution f(x,y) = integral M(x,y)dx + g(y) = C
Test for exactness
M partial y = n partial x
Homogenous Equations
If the expression can have x/y or y/x or some factor of them or constants then it’s homogenous. Every x must have a y and every y must have a x
Solving homogenous equations
- Substitute y/x with v
Now the de should be dy/dx = G(v) - we need to make dy/dx in terms of v and x
v = y/x so y=vx and dy/dx = x (dv/dx) + v - sub this into separate variable and integrate
- replace v with its respected terms of x and y
Bernoulli Equations
When an equation looks linear but y is raised to a power
dy/dx + P(x) y = Q(x) y^n when n > 1
Solving Bernoulli equations
- note value of n
- note value of 1-n
- v = y^(1-n)
- dv/dx = (1-n)y^n
- use magic equaiton
dV/dx + (1-n) P(x) v = (1-n)Q(x)
