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Thread: Dif EQ

  1. #1
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    Dif EQ

    Another Dif EQ, yay.

    I have to solve the following Dif EQ, and give the interval for which the largest general solution is defined:

    dP/dt + 2tP = P + 4t - 2

    So I don't think I can do separation of variables, so I tried integrating factors...

    P(x) = 2t

    M(x) = e^(int(2t)dt) = e^(2t^3/3)

    So multiply M(x) to each term:

    e^(2t^3/3)*dP/dt + e^(2t^3/3)*2tP = e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3)

    So this looks very messy now...

    Again, it's the part after this that I get stuck.
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    Upon re-examining this,

    Would it be:

    int((e^(2t^3/3)*P)') = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))

    => (e^(2t^3/3)*P) = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    Another Dif EQ, yay.

    I have to solve the following Dif EQ, and give the interval for which the largest general solution is defined:

    dP/dt + 2tP = P + 4t - 2

    So I don't think I can do separation of variables, so I tried integrating factors...

    P(x) = 2t
    you are wrong since this line

    Quote Originally Posted by Ideasman View Post
    Upon re-examining this,

    Would it be:

    int((e^(2t^3/3)*P)') = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))

    => (e^(2t^3/3)*P) = int(e^(2t^3/3)*P + e^(2t^3/3)*4t - 2*e^(2t^3/3))
    thus, this is wrong as well
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    Quote Originally Posted by Jhevon View Post
    you are wrong since this line
    But why? I thought you took the stuff that wasn't on the dependent variable and let that equal P(x). Is it because of the RHS of the equation? Do I need to move things first? Suppose it was just

    dP/dt + 2tP = 0. Would P(x) be 2t then?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    But why? I thought you took the stuff that wasn't on the dependent variable and let that equal P(x). Is it because of the RHS of the equation? Do I need to move things first?
    you had $\displaystyle P' + 2tP = P + 4t - 2$

    $\displaystyle \Rightarrow P' + 2tP - P = 4t - 2$

    $\displaystyle \Rightarrow P' + (2t - 1)P = 4t - 2$

    now what do you think $\displaystyle P(x)$ is?

    (in this case, using "P" gets confusing, because that's the variable we're working with in the first place, but oh well)

    Suppose it was just

    dP/dt + 2tP = 0. Would P(x) be 2t then?
    yes, but you would not use the integrating factor in this case, this is a homogeneous first order DE, it is separable
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    Quote Originally Posted by Jhevon View Post
    you had $\displaystyle P' + 2tP = P + 4t - 2$

    $\displaystyle \Rightarrow P' + 2tP - P = 4t - 2$

    $\displaystyle \Rightarrow P' + (2t - 1)P = 4t - 2$

    now what do you think $\displaystyle P(x)$ is?
    2t - 1

    Okay, so if P(x) = 2t - 1

    That means M(x) = e^(int(2t - 1)dt) = e^(t^2 - t)

    Multiply M(x) by each term:

    e^(t^2 - t)*P' + e^(t^2 - t)*P = e^(t^2 - t)*4t - 2*e^(t^2 - t)

    Okay:

    int[e^(t^2 - t)*P)'] = int(e^(t^2 - t)*4t - 2*e^(t^2 - t))

    => e^(t^2 - t)*P = {Since this can't be integrated, I think I've done something wrong }
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    2t - 1

    Okay, so if P(x) = 2t - 1

    That means M(x) = e^(int(2t - 1)dt) = e^(t^2 - t)

    Multiply M(x) by each term:

    e^(t^2 - t)*P' + e^(t^2 - t)*P = e^(t^2 - t)*4t - 2*e^(t^2 - t)

    Okay:

    int[e^(t^2 - t)*P)'] = int(e^(t^2 - t)*4t - 2*e^(t^2 - t))

    => e^(t^2 - t)*P = {Since this can't be integrated, I think I've done something wrong }
    no, you're correct. new strategy: let's go for separable

    $\displaystyle P' + (2t - 1)P = 4t - 2$

    $\displaystyle \Rightarrow P' = 4t - 2 - (2t - 1)P$

    $\displaystyle \Rightarrow P' = 2(2t - 1) - (2t - 1)P$

    $\displaystyle \Rightarrow P' = (2t - 1)(2 - P)$

    $\displaystyle \Rightarrow \frac {P'}{2 - P} = 2t - 1$

    now continue
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    Quote Originally Posted by Jhevon View Post
    no, you're correct. new strategy: let's go for separable

    $\displaystyle P' + (2t - 1)P = 4t - 2$

    $\displaystyle \Rightarrow P' = 4t - 2 - (2t - 1)P$

    $\displaystyle \Rightarrow P' = 2(2t - 1) - (2t - 1)P$

    $\displaystyle \Rightarrow P' = (2t - 1)(2 - P)$

    $\displaystyle \Rightarrow \frac {P'}{2 - P} = 2t - 1$

    now continue
    Ah, tricky algebra manipulation.

    Okay, so from where you left off:

    dP/dt * 1/(2-P) = 2t - 1

    => dP/(2-P) = (2t-1)dt

    Take integral of both sides:

    -ln|P-2| = t^2 - t + C

    Now the second part of the question asked to find the largest interval over which the GENERAL sol'n is defined. This looks defined everywhere, except for when the ln is 0 or negative, but since we have the absolute signs, we don't have to worry about it being neg. Therefore, it's defined (-infinity, 2) or (2, infinity). How am I supposed to know which it's in though.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    Ah, tricky algebra manipulation.

    Okay, so from where you left off:

    dP/dt * 1/(2-P) = 2t - 1

    => dP/(2-P) = (2t-1)dt

    Take integral of both sides:

    -ln|P-2| = t^2 - t + C

    Now the second part of the question asked to find the largest interval over which the GENERAL sol'n is defined. This looks defined everywhere, except for when the ln is 0 or negative, but since we have the absolute signs, we don't have to worry about it being neg. Therefore, it's defined (-infinity, 2) or (2, infinity). How am I supposed to know which it's in though.
    read, again, what i said here regarding this question
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    Quote Originally Posted by Jhevon View Post
    read, again, what i said here regarding this question
    That's the whole point; I don't have a point t_0.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    That's the whole point; I don't have a point t_0.
    what is P(x)? and where is it continuous?
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