Question

find general solution dy/dt+2y=0

Answer 1

try to separate the variables

Answer 2

and then?

Answer 3

integrate to find the solution

Answer 4

when separate the variables would it be dy+2y=dt

Answer 5

the equation you gave is not equivalent to one you have produced just now

Answer 6

how does the equation look after separation of variables?

Answer 7

you should we able to write the one you have posted as f(y) dy =g(x) dx

Answer 8

subtract 2y on both sides

Answer 9

\[\frac{dy}{dt}=-2y \]

Answer 10

do you know where to go from here?

Answer 11

you still are trying to separate the variables

Answer 12

is it -1/2y (dy)=dt

Answer 13

yes you divide both sides by -2y and multiply both sides by dt
\[\frac{dy}{-2y}=dt \]
this is in the form of f(y) dy=g(t) dt

this is a separation by variables type of problem

now just integrate both sides

Answer 14

is it \[\ln \left| -2y \right|=t+c\]

Answer 15

not exactly

Answer 16

you should try integrating -1/(2y) w.r.t y again

Answer 17

\[\int\limits_{}^{}\frac{-1}{2y} dy =\frac{-1}{2}\int\limits_{}^{}\frac{1}{y} dy=\frac{-1}{2}\ln|y|-c \neq \ln|-2y|-c\]

Answer 18

why is it -c

Answer 19

you always do plus some constant when you have a indefinite integral

Answer 20

it is just the general antiderivative

Answer 21

but in this case is - not +
that's what you put -c

Answer 22

you don't know if c is negative or positive
it doesn't matter if we put +3*c
or -5000*c
or -c/4
or +c/4

Answer 23

ok then so the answer is -1/2lny-c=t+c

Answer 24

those constants don't exactly have to be the same value

you only need need put C on onside anyways since a difference or sum of constants is a constant anyhow

the problem i had with your "answer" is that you didn't integrate -1/(2y) correctly as I said above

Answer 25

\[\text{ so yeah we can say we have } \frac{-1}{2}\ln|y|-c=t+C\]
then adding little c on both sides we have
\[\frac{-1}{2}\ln|y|=t+C+c\]
but again a constant +a constant is still a constant and I will call the sum of those constants K
\[\frac{-1}{2}\ln|y|=t+K\]

Answer 26

you can solve that for y if you want

Answer 27

how can I solve for y?

Answer 28

by isolating it on one side

Answer 29

does it cancel with e?

Answer 30

are you asking me if y=ln(x) and y=e^x are inverse functions?
if so yeah.
this means
\[e^{ln|y|}=y\]
\text{ or }
\[\ln(e^y)=y\]

you only need that first one there though

be careful you need to do something with constant multiple on the natural log part (speaking of the -1/2)

Answer 31

do i move -1/2 to the other side?

Answer 32

well you don't just move -1/2 on both sides
you multiply -2 on both sides

Answer 33

You might need to go back and do a review of algebra, like on solving linear equations and solving equations involving log and the exponential expressions.

Answer 34

can you show me how can i do it, like solve for y?, please

Answer 35

have you already multiplied both sides by -2?

Answer 36

yea

Answer 37

and what did you get?

Answer 38

\[y=e ^{t+c(-2)}\]

Answer 39

ok you had \[\frac{-1}{2}\ln|y|=t+K \]
multiplying -2 on both sides gives
\[\ln|y|=-2t+C\]

Answer 40

-2*K is still just a constant

Answer 41

now use the thing thing I mention about inverses to rewrite this so you have y isolated

Answer 42

\[y=e ^{-2t+c}\]

Answer 43

that is groovy
now another fun way people like to write that is:
\[y=e^{-2t+c}=e^{-2t}e^c=e^{-2t}C=Ce^{-2t}\]

Answer 44

since e^c is again just a constant
a positive constant but still a constant

Answer 45

ok thanks