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

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- anonymous

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

- jamiebookeater

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- jtvatsim

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

- jtvatsim

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

- anonymous

to test the equation to see if its true

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- anonymous

- jtvatsim

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}.

- jtvatsim

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. :)

- anonymous

i dont get it :(

- jtvatsim

OK, can you take the derivative of y?

- UsukiDoll

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

- welshfella

I'm straining my memory now as its been a long time. I think this a first order linear equation
dy/dx - 3y = -3

- welshfella

Integrating factor = e ^ [INT -3dx) = e^-3x
next you multiply each term in the equation by this integrating factor

- welshfella

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

- UsukiDoll

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.

- UsukiDoll

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

- welshfella

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

- welshfella

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

- welshfella

now we divide by e^-3x
y = 1 + C / e^-3x
not sure about that!

- welshfella

that gives y = ce^3x + 1
different sign!!???

- UsukiDoll

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

- welshfella

i might have gone wrong somewhere...

- welshfella

- its about 50 years since I last did these lol

- welshfella

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

- welshfella

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

- UsukiDoll

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

- mathmate

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
\(y+1=Ce^{3(y+1)}\)
\(y=Ce^{3(y+1)}-1\)

- UsukiDoll

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.

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