Question

does anyone understand Differential equations

Answer 1

yes

Answer 2

einstein did i think

Answer 3

newton faked it lol

Answer 4

lol

Answer 5

i just begun D.E. and i am totally lost.
find a function, y=f(x) satisfying the given differential equation and the prescribed initial condition.
dy/dx= 2x+1; y(0)=3
where do i start

Answer 6

with integration id assume

Answer 7

multiply both sides by dx .. then integrate both sides

Answer 8

\[y' = 2x+1\]
\[\int y' = \int 2x+1\]
\[y=x^2+x +C\]

Answer 9

plug in x=0 and y=3 to solve for C

Answer 10

this is a seperable DE

Answer 11

yeah, semantics :)

Answer 12

:D:D

Answer 13

you do these just like seperable
this is 2 sections before seperable

Answer 14

Initial Value problem?

Answer 15

intergrals as general and particular solutions

Answer 16

forget what i said, its done the way amistre showd u

Answer 17

yay!!

Answer 18

hehehe

Answer 19

ok appreciate it!

Answer 20

\[Y(x)=\int y'\ dx\]
\[Y(x)=\int (2x+1)dx\]
stuff like that

Answer 21

it just shows you the equation and intial condition and you have to choose what to do. So when you work it out it becomes what you are doing with the integral

Answer 22

\[\frac{dy}{dx}=2x+1\]\[dy=(2x+1)dx\]now integrate both sides\[\int\limits{dy}=\int\limits{(2x+1)dx}\]\[y=x^2+x+c\]

Answer 23

yea, just like that, thanks,

Answer 24

:)

Answer 25

i need help with intergals, it has been 2 yrs since i have done them. x times sq. root of x^2+9

Answer 26

let u=x^2+9
du=2x

so integration becomes\[\frac{1}{2} \int\limits{ \sqrt u}du\]

Answer 27

integrate sqrt u then substitute the x back

Answer 28

du=2xdx **

Answer 29

I am really sorry, i have to leave now. Post a new question and someone else could help you

Answer 30

thanks

Answer 31

is \[^{?} \int\limits_{?}^{?} xe ^{-x}\]
u substitution or by parts integral

Answer 32

or simply recognize that your missing a -1 in front

Answer 33

i missed read it

Answer 34

by parts with that one

Answer 35

alright

Answer 36

e^-x
+ x -e^-x |
-1 e^-x | = x e^-x - e^-x
+0 -----

Answer 37

+C if youaint got bounds

Answer 38

i missed a negative on there .... bad fingers, bad!!

Answer 39

\[-xe^{-x}-e^{-x}\ or\ -e^{-x}(x+1)\]