Question

dy/dx=-3y-2; y(2) = 0 ... when solved the equation i have got y= -2/3 +Ke^-3t now what shall i do with y(2) = 0...pls help me

Answer 1

you plug in x=2 in the equation you got to find the value of k

Answer 2

and put y=0

Answer 3

i didnt understand , you mean in the question
pls explain

Answer 4

not in the question, the equation u got
y= -2/3 +Ke^-3t

y(2) = 0
means , when t=2, y=0

so plug in t=2,y=0 in y= -2/3 +Ke^-3t

Answer 5

your independent variable is "x" right ?
so you should have got

y= -2/3 +Ke^-3x

isn't it ??
hence
y(2) = 0
means , when x=2, y=0

so plug in x=2,y=0 in y= -2/3 +Ke^-3x

Answer 6

got it ?

Answer 7

yes.. thanks... let me try

Answer 8

k, take your time :)

Answer 9

now 0=-2/3+Ke^-6 am i right..

Answer 10

yes, try to isolate k now

Answer 11

ok... i am trying...:)

Answer 12

that's right you almost there,

Answer 13

when you get K,
just plug it into

y(x) = -2/3 +Ke^(-3x)

and simplify a little bit,

and that's it.

Answer 14

i lost it .. i tried k=-2/3+e^-6 if its right ..then got stuck again

Answer 15

you still had to isolate k from -2/3+ke^-6=0
start by adding 2/3 on both sides

Answer 16

you had

0=-2/3+Ke^-6

add 2/3 to both sides and multiply by e^6

Answer 17

i am not reaching anywhere.... pls show me this then i will get an idea

Answer 18

0=-2/3+Ke^-6

2/3 = ke^{-6}

2/3 * e^6 = k e^{-6} e^6

so, k= 2/3e^6

Answer 19

in the third line u used exponential rule.. like m^n* m^m = m^n+m then that became 1 am i right ...

Answer 20

yes, exactly

-6+6=0
e^0=1

i multiplied by e^6 to make the exponent =0

Answer 21

ok now its clear for me.. wow my brain so relaxed.. a million thanks will come back soon with an another question :)

Answer 22

you are most welcome ^_^
oh, and bdw
\[ \begin{array}l\color{red}{\text{W}}\color{orange}{\text{E}}\color{#e6e600}{\text{L}}\color{green}{\text{C}}\color{blue}{\text{O}}\color{purple}{\text{M}}\color{purple}{\text{E}}\color{red}{\text{ }}\color{orange}{\text{t}}\color{#e6e600}{\text{o}}\color{green}{\text{ }}\color{blue}{\text{O}}\color{purple}{\text{P}}\color{purple}{\text{E}}\color{red}{\text{N}}\color{orange}{\text{S}}\color{#e6e600}{\text{T}}\color{green}{\text{U}}\color{blue}{\text{D}}\color{purple}{\text{Y}}\color{purple}{\text{!}}\color{red}{\text{!}}\color{orange}{\text{}}\end{array} \]