Turn on thread page Beta
    • Thread Starter
    Offline

    20
    ReputationRep:
    is it possible to solve

    \frac{dy}{dt}  = \sqrt y - \sqrt x ?

    the bear

    :badger:
    • Thread Starter
    Offline

    20
    ReputationRep:
    i have plotted a gradient field using x and y axes...

    how should the solution curves from the field be used if the formula is dy/dt = x^0.5 - y^0.5 rather than dy/dx = x^0.5 - y^0.5 ?

    the baer

    :confused::confused:
    Offline

    0
    ReputationRep:
    How about we re-arrange it to: \frac{dy}{dt} - \sqrt y = - \sqrt x

    Notice how the differential equation involves only a "dy" and "dt" NOT both "dy" and "dx" so we can simply treat \sqrt x as a constant

    Now it's in the form \frac{dy}{dt} - \sqrt y = constant so we could perhaps solve it using the Integrating Factor

    I've not actually tried it out on paper mind, so it may not work. But hope it helps
    Offline

    17
    ReputationRep:
    (Original post by McMannBean)
    How about we re-arrange it to: \frac{dy}{dt} - \sqrt y = - \sqrt x

    Notice how the differential equation involves only a "dy" and "dt" NOT both "dy" and "dx" so we can simply treat \sqrt x as a constant

    Now it's in the form \frac{dy}{dt} - \sqrt y = constant so we could perhaps solve it using the Integrating Factor

    I've not actually tried it out on paper mind, so it may not work. But hope it helps
    You can't use the integrating factor method because the ODE is non-linear.
    Offline

    0
    ReputationRep:
    let's try a substitution then:

    Let U = sqrt(y)

    so we end up with 2U*dU/dt - U = const

    => dU/dt - const*U^-1 = 1

    This should be solveable
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by McMannBean)
    let's try a substitution then:

    Let U = sqrt(y)

    so we end up with 2U*dU/dt - U = const

    => dU/dt - const*U^-1 = 1

    This should be solveable
    How do we know that x is independent of t? [If it's not then we certainly can't assume it's constant.]

    I think we lack enough information to be able to find a solution here. Is there more to the question?
    • Thread Starter
    Offline

    20
    ReputationRep:
    Thanks so much guys for those helpful suggestions. As Nuodai said it would be better if x were a function of t .

    it may be that the function is quadratic:

    x = ( k - t )^2

    where k is a constant

    then the equation would become:

    dy/dt = y^0.5 - ( k - t )

    & then the tangent field with y and t axes would be appropriate: here the horizontal axis is t: k is 20
    Attached Images
     
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 15, 2011
Poll
Is the Big Bang theory correct?
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.