anonymous
  • anonymous
GENERAL SOLUTION y'=3(y+1) ANS: y=ce^3x-1
Mathematics
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anonymous
  • anonymous
GENERAL SOLUTION y'=3(y+1) ANS: y=ce^3x-1
Mathematics
jamiebookeater
  • jamiebookeater
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jtvatsim
  • jtvatsim
The question is: given the differential equation, find the general solution?
jtvatsim
  • jtvatsim
Or are you supposed to test the equation to see if its true?
anonymous
  • anonymous
to test the equation to see if its true

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anonymous
  • anonymous
jtvatsim
  • 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
  • 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
  • anonymous
i dont get it :(
jtvatsim
  • jtvatsim
OK, can you take the derivative of y?
UsukiDoll
  • 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
  • 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
  • welshfella
Integrating factor = e ^ [INT -3dx) = e^-3x next you multiply each term in the equation by this integrating factor
welshfella
  • welshfella
I'll have to refresh my memory on this stuff....
UsukiDoll
  • 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
  • UsukiDoll
either separation of variables or integrating factor works... However, separation of variables seems to be the fastest way to retrieve the solution
welshfella
  • welshfella
e^-3x * dy/dx - 3y e^-3x = -3 e^-3x y e^-3x = INT -3 -3x dx
welshfella
  • welshfella
y e^-3x = e^-3x + C
welshfella
  • welshfella
now we divide by e^-3x y = 1 + C / e^-3x not sure about that!
welshfella
  • welshfella
that gives y = ce^3x + 1 different sign!!???
UsukiDoll
  • UsukiDoll
maybe it's a typo? I'll do the separation of variables way and pm you what I got @welshfella
welshfella
  • welshfella
i might have gone wrong somewhere...
welshfella
  • welshfella
- its about 50 years since I last did these lol
welshfella
  • welshfella
oh I see now i made an error right at the beginning its dy/dx - 3y = 3 ( not -3)
welshfella
  • welshfella
it is -1 not 1 y = ce^3x - 1
UsukiDoll
  • UsukiDoll
yup and also you have to take that minus sign into consideration. so your p(x) for integrating factor is -3
mathmate
  • 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
  • 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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