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