Turn on thread page Beta
    • Thread Starter
    Offline

    5
    ReputationRep:
    Using the substitution y=1/u, show that the general solution to (1/y^2)dy/dx + 1/y = e^(2x) is y=1/(Ae^x - e^(2x)???

    What I’ve got so far:
    dy/du=-1/u^2
    u^2(dy/dx) + u = e^(2x)
    -du/dx + u = e^(2x)
    -du + u dx = e^(2x) dx
    But I’m stuck here because you cannot integrate (u dx).

    Many thanks
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    15
    ReputationRep:
    (Original post by Constan4699)
    Using the substitution y=1/u, show that the general solution to (1/y^2)dy/dx + 1/y = e^(2x) is y=1/(Ae^x - e^(2x)???

    What I’ve got so far:
    dy/du=-1/u^2
    u^2(dy/dx) + u = e^(2x)
    -du/dx + u = e^(2x)
    -du + u dx = e^(2x) dx
    But I’m stuck here because you cannot integrate (u dx).

    Many thanks
    It looks as though you have just multiplied by dx which you can't do.
    You will need to solve the resulting ODE by using an integrating factor because this equation is not separable.
    Offline

    10
    ReputationRep:
    (Original post by Constan4699)
    Using the substitution y=1/u, show that the general solution to (1/y^2)dy/dx + 1/y = e^(2x) is y=1/(Ae^x - e^(2x)???

    What I’ve got so far:
    dy/du=-1/u^2
    u^2(dy/dx) + u = e^(2x)
    -du/dx + u = e^(2x)
    -du + u dx = e^(2x) dx
    But I’m stuck here because you cannot integrate (u dx).

    Many thanks
    Consider your third line:
     \frac{dy}{dx} - u=-e^{2x}

    Have you covered the use of integrating factors? If so do you see how to apply one here?
    Offline

    18
    ReputationRep:
    Do you know how to do first order nonseperable ODEs?

    Look at the coefficient of the u term to determine the integrating factor.

    A little prod,

    Spoiler:
    Show
    does e^{\int P \mathrm dx} look familiar?
    • Thread Starter
    Offline

    5
    ReputationRep:
    No I’ve never seen the use of integrating factors - probably why I’m unable to progress then. Please help?
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    18
    ReputationRep:
    (Original post by Constan4699)
    No I’ve never seen the use of integrating factors - probably why I’m unable to progress then. Please help?
    Maybe watch a video (KhanAcademy is normally good) on integrating factors before continuing? Probably better than explaining it to you here, on the spot.
    • Thread Starter
    Offline

    5
    ReputationRep:
    So from my understanding it seems that the integrating factor is e^(-x) (hope this is correct?). So starting again:

    du/dx - u = -e^(2x)
    So e^(-x)(du/dx) - ue^(-x) = - e^x
    Can anyone give me any pointers from here?
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    18
    ReputationRep:
    (Original post by Constan4699)
    So from my understanding it seems that the integrating factor is e^(-x) (hope this is correct?). So starting again:

    du/dx - u = -e^(2x)
    So e^(-x)(du/dx) - ue^(-x) = - e^x
    Can anyone give me any pointers from here?
    Recall the product rule,

    (uv)' = u'v + uv'

    Also that \displaystyle \frac{\mathrm d}{\mathrm dx} u = \frac{\mathrm du}{\mathrm dx}. (you should also be familiar with the derivative of e^x) Does this help?
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by _gcx)
    Recall the product rule,

    (uv)' = u'v + uv'

    Also that \displaystyle \frac{\mathrm d}{\mathrm dx} u = \frac{\mathrm du}{\mathrm dx}. (you should also be familiar with the derivative of e^x) Does this help?
    Yep thank you very much I think I’ve got it now
    Posted on the TSR App. Download from Apple or Google Play
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 10, 2017
Poll
Do you think parents should charge rent?
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.