Question

GENERAL SOLUTION
y'=3(y+1) ANS: y=ce^3x-1

Answer 1

The question is: given the differential equation, find the general solution?

Answer 2

Or are you supposed to test the equation to see if its true?

Answer 3

to test the equation to see if its true

Answer 4

@jtvatsim

Answer 5

OK, we want to plug it into the differential equation
y' = 3(y + 1)
and check that both sides are equal.

We already know that what y equals, so we can evaluate the right-hand side (RHS) of the equation:

RHS
3(y + 1) = 3([ce^{3x} -1] + 1) = 3(ce^{3x}) = 3ce^{3x}.

Answer 6

Is this the same as the left-hand side (LHS)? Well, we need to find the value of
y'

We can just take the derivative of our given y to find this. See what you get. :)

Answer 7

i dont get it :(

Answer 8

OK, can you take the derivative of y?

Answer 9

I think this question means we need to verify that the answer is y=e^{3x}-1 using ode techniques.
use separation of variables

Answer 10

I'm straining my memory now as its been a long time. I think this a first order linear equation

dy/dx - 3y = -3

Answer 11

Integrating factor = e ^ [INT -3dx) = e^-3x

next you multiply each term in the equation by this integrating factor

Answer 12

I'll have to refresh my memory on this stuff....

Answer 13

yeah we can use integrating factor
Actually there is more than one method to solving this equation and it should lead to the same answer.

Answer 14

either separation of variables or integrating factor works...
However, separation of variables seems to be the fastest way to retrieve the solution

Answer 15

e^-3x * dy/dx - 3y e^-3x = -3 e^-3x

y e^-3x = INT -3 -3x dx

Answer 16

y e^-3x = e^-3x + C

Answer 17

now we divide by e^-3x
y = 1 + C / e^-3x

not sure about that!

Answer 18

that gives y = ce^3x + 1


different sign!!???

Answer 19

maybe it's a typo? I'll do the separation of variables way and pm you what I got @welshfella

Answer 20

i might have gone wrong somewhere...

Answer 21

- its about 50 years since I last did these lol

Answer 22

oh I see now i made an error right at the beginning
its dy/dx - 3y = 3 ( not -3)

Answer 23

it is -1 not 1
y = ce^3x - 1

Answer 24

yup and also you have to take that minus sign into consideration. so your p(x) for integrating factor is -3

Answer 25

How about try substitution:
with p=y+1
then y=p-1, y'=p'
y'=3(y+1) becomes
p'=3(p)
p'/p=3
integrate both sides
log(p)=3p+C
raise to power of e
\(p=e^{3p+C}=Ce^{3p}\) [note the two C's have different values]
Back substitute
\(y+1=Ce^{3(y+1)}\)
\(y=Ce^{3(y+1)}-1\)

Answer 26

substitution method is a killer torture method for this problem though ...like it works but I still think that separation of variables is the fastest way to get the solution.