Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:


    I get the first equation down to:

    |P|=Ae^{-\alpha t}

    How do I get rid of the mod signs (something to do with plus/minus?) and apply the initial conditions to find A?

    Then how would I go about solving equation 2?
    Offline

    12
    ReputationRep:
    Can P be negative?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by JoMo1)
    Can P be negative?
    I suppose not from the context as it is an 'element' - a bit vague though.

    So that would give A=P_0

    which means P=P_0 e^{-\alpha t}

    How would I solve

    \frac{dQ}{dt}=\alpha P - \beta Q ?
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    I suppose not from the context as it is an 'element' - a bit vague though.

    So that would give A=P_0

    which means P=P_0 e^{-\alpha t}

    How would I solve

    \frac{dQ}{dt}=\alpha P - \beta Q ?
    use integrating factors
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by thebadgeroverlord)
    use integrating factors
    I haven't learnt that yet so I can't do it but just in general, how would I get rid of a mod sign?

    Would it have become:

    P=\pm Ae^{-\alpha t}

    A=\pm P_0 ?
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    I haven't learnt that yet so I can't do it but just in general, how would I get rid of a mod sign?

    Would it have become:

    P=\pm Ae^{-\alpha t}

    A=\pm P_0 ?
    just absorb the signs into the constant A then applying boundary intial conidtions will determine it.
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    I haven't learnt that yet so I can't do it but just in general, how would I get rid of a mod sign?

    Would it have become:

    P=\pm Ae^{-\alpha t}

    A=\pm P_0 ?
    take bQ to LHS

    to solve the equation multiply it through by u(x) so that

    LHS=w dQ/dt + bwQ

    then look for w such that LHS=d(wQ)/dt

    so use product rule and compare with wdQ/dt + bwQ to get a DE for w and solve that.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by thebadgeroverlord)
    take bQ to LHS

    to solve the equation multiply it through by u(x) so that

    LHS=w dQ/dt + bwQ

    then look for w such that LHS=d(wQ)/dt

    so use product rule and compare with wdQ/dt + bwQ to get a DE for w and solve that.
    So w\frac{dQ}{dt}+\beta wQ=Q\frac{dw}{dt}+w\frac{dQ}{dt}

    \beta w=\frac{dw}{dt}

    \int \beta dt=\int \frac{1}{w} dw

    ln|w|=\beta t+c

    |w|=Ae^{\beta t}

    where to now?
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    So w\frac{dQ}{dt}+\beta wQ=Q\frac{dw}{dt}+w\frac{dQ}{dt}

    \beta w=\frac{dw}{dt}

    \int \beta dt=\int \frac{1}{w} dw

    ln|w|=\beta t+c

    |w|=Ae^{\beta t}

    where to now?
    You can take A = 1 so w=exp(bt)
    you don't need to bother with modulus signs for solns of DEs.

    so LHS = d(wQ)/dt
    RHS = wP

    so d(wQ)/dt = wP so integrate both sides wrt t.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by thebadgeroverlord)
    You can take A = 1 so w=exp(bt)
    you don't need to bother with modulus signs for solns of DEs.

    so LHS = d(wQ)/dt
    RHS = wP

    so d(wQ)/dt = wP so integrate both sides wrt t.
    Where's RHS = wP come from?
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    Where's RHS = wP come from?
    dQ/dt = -bQ - dP/dt

    but dP/dt = -aP

    => dQ/dt + bQ = aP

    and then you multiplied the equation by w. Sorry I missed out the a in my last post.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by thebadgeroverlord)
    dQ/dt = -bQ - dP/dt

    but dP/dt = -aP

    => dQ/dt + bQ = aP

    and then you multiplied the equation by w. Sorry I missed out the a in my last post.
    So  \frac{d}{dt}(wQ)=w\alpha P

    \Rightarrow \frac{d}{dt}(wQ)=w\alpha P_0 e^{-\alpha t}

    wQ=-\frac{w\alpha}{\alpha}P_0 e^{-\alpha t}

    Q=-P_0 e^{-\alpha t}

    EDIT: This is wrong as w is a function not a constant isn't it?
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    So  \frac{d}{dt}(wQ)=w\alpha P

    \Rightarrow \frac{d}{dt}(wQ)=w\alpha P_0 e^{-\alpha t}

    wQ=-\frac{w\alpha}{\alpha}P_0 e^{-\alpha t}

    Q=-P_0 e^{-\alpha t}

    EDIT: This is wrong as w is a function not a constant isn't it?
    When you integrate the RHS, w is also a function of t/
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by thebadgeroverlord)
    When you integrate the RHS, w is also a function of t/
    so does it have to be done by parts?

    |f(y)-f(x)|\leq |y-x||x^2+xy+y^2|

    and assume y\neq x .

    Then |\frac{f(y)-f(x)}{y-x}|\leq |x^2+xy+y^2| .

    For x=0 we then have |\frac{f(y)-f(0)}{y}|\leq |y^2| .

    Now let y \rightarrow 0 . Then the RHS converges to 0, implying that f is differentiable at 0 and f'(0)=0 .

    --------

    f is differentiable on (0,\infty) if the following limit exists for all fixed x on the interval (0,\infty)

    \displaystyle \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x)

    With f(x)=\sqrt{x},\;f(x+h)=\sqrt{x+h  } for x\in (0,\infty) :

    f'(x)=\displaystyle \lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}

    =\displaystyle \lim_{h\to 0} \left( \frac{\sqrt{x+h}-\sqrt{x}}{h} \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt  {x+h}+\sqrt{x}} \right)

    =\displaystyle \lim_{h\to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{\sqrt{x}+\sqrt{x}} = \frac{1}{2\sqrt{x}}

    =|\displaystyle \lim_{y\to x} \frac{f(y)-f(x)}{y-x}| = |f'(x)|\leq |\lim_{y\to x} x^2+xy+y^2|=|x^2+x^2+x^2|=|3x^2|  =3x^2
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    so does it have to be done by parts?
    combine the powers of e
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by thebadgeroverlord)
    combine the powers of e
    So \frac{d}{dw}(wQ)=\alpha P_0 e^{-\alpha t}e^{\beta t}

    \frac{d}{dw}(wQ)= \alpha P_0 e^{(\beta -\alpha)t}

    wQ=\frac{\alpha P_0}{\beta -\alpha}e^{(\beta -\alpha)t}

    Qe^{\beta t}=\frac{\alpha P_0}{\beta -\alpha}e^{\beta t}e^{-\alpha t}

    Q=\frac{\alpha P_0}{\beta -\alpha}e^{-\alpha t}
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    So \frac{d}{dw}(wQ)=\alpha P_0 e^{-\alpha t}e^{\beta t}

    \frac{d}{dw}(wQ)= \alpha P_0 e^{(\beta -\alpha)t}

    wQ=\frac{\alpha P_0}{\beta -\alpha}e^{(\beta -\alpha)t}

    Qe^{\beta t}=\frac{\alpha P_0}{\beta -\alpha}e^{\beta t}e^{-\alpha t}

    Q=\frac{\alpha P_0}{\beta -\alpha}e^{-\alpha t}
    That looks right to me.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by thebadgeroverlord)
    That looks right to me.
    But applying the initial condition Q(0)=0 that would imply

    \frac{\alpha P_0}{\beta -\alpha}=0

    Wouldn't this imply that P_0=0 which can't be right as P has to initially be positive as it is "decaying" otherwise the system won't work?
    Offline

    2
    ReputationRep:
    (Original post by TheEd)
    But applying the initial condition Q(0)=0 that would imply

    \frac{\alpha P_0}{\beta -\alpha}=0

    Wouldn't this imply that P_0=0 which can't be right as P has to initially be positive as it is "decaying" otherwise the system won't work?
    Just gone through it all myself and realised you've missed a constant of integration when you integrated awP so you should have an extra Cexp(-Bt) on the RHS of your soln
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by thebadgeroverlord)
    Just gone through it all myself and realised you've missed a constant of integration when you integrated awP so you should have an extra Cexp(-Bt) on the RHS of your soln
    Ah yes it's because I didn't put in explicit integral signs so I forgot!

    So it should be:

    Q(t)=\frac{\alpha P_0}{\beta -\alpha}(e^{-\alpha t}-e^{-\beta t}) provided \alpha \neq \beta
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.