Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Solve \frac{dy}{dx}=\frac{2y^4+x^4}{xy  ^3}

    Let z=\frac{y}{x}

    \Rightarrow y=xz \Rightarrow \frac{dy}{dx}=x\frac{dz}{dx}+z

    Now the differential equation can be written as \frac{dy}{dx}=\frac{2y^4}{xy^3}+  \frac{x^4}{xy^3} which simplifies to 2(\frac{y}{x})+({\frac{x}{y}})^3  =2z+z^{-3}

    So x\frac{dz}{dx}+z=2z+z^{-3}

    x\frac{dz}{dx}=z+z^{-3}

    where to go from here?
    Offline

    20
    Divide across, then integrate, by the looks of things.
    Offline

    9
    ReputationRep:
    separate variables
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by + polarity -)
    Divide across, then integrate, by the looks of things.
    So \int\frac{1}{x}dx=\int\frac{1}{z  +z^{-3}}dz

    How to integrate RHS?
    Offline

    9
    ReputationRep:
    well RHS is same as z^3/(z^4 +1) dz
    the top is close to the derivative of the bottom
    Offline

    20
    (Original post by TheEd)
    So \int\frac{1}{x}dx=\int\frac{1}{z  +z^{-3}}dz

    How to integrate RHS?
    Would it become... \int\frac{z^{3}}{z^{4}+1}dz? Which = \frac{1}{4}\int\frac{4z^{3}}{z^{  4}+1}dz

    Ah! Use the trick MrGreedy alluded to!
    Offline

    0
    ReputationRep:
    (Original post by + polarity -)
    Would it become... \int\frac{z^{3}}{z^{4}+1}dz?

    Ouch. You could try partial fractions, but I have no idea if it would work. :sigh:
    :nope: it's \frac{1}{4}ln(z^4+1)
    Offline

    9
    ReputationRep:
    (Original post by + polarity -)
    Would it become... \int\frac{z^{3}}{z^{4}+1}dz?

    Ah! Use the trick MrGreedy alluded to!
    all you need to know is that
    (integral sign) f'(z)/f(z) dz = ln(f(z)) + c
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by MrGreedy)
    well RHS is same as z^3/(z^4 +1) dz
    the top is close to the derivative of the bottom
    So \frac{1}{4}ln|z^4+1|+c ?
    Offline

    9
    ReputationRep:
    (Original post by TheEd)
    So \frac{1}{4}ln|z^4+1|+c ?
    yes
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Pheylan)
    :nope: it's \frac{1}{4}ln(z^4+1)
    Yes that's what I got when MrGreedy reminded me of the derivative trick
    Offline

    20
    (Original post by Pheylan)
    :nope: it's \frac{1}{4}ln(z^4+1)
    I edited my post. :p:
    Offline

    0
    ReputationRep:
    (Original post by TheEd)
    Yes that's what I got when MrGreedy reminded me of the derivative trick
    i got to the \frac{z^3}{z^4+1} stage but i forgot the trick myself :o:
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by MrGreedy)
    yes
    So \frac{1}{x}+c=\frac{1}{4}ln|z^4+  1|

    Ae^{\frac{4}{x}}=|z^4+1|

    I get y=\sqrt[4]{x^4(Ae^{\frac{4}{x}}-1})

    The question also tells me to verify the solution. How would I verify if this is correct?
    Offline

    9
    ReputationRep:
    (Original post by TheEd)
    So \frac{1}{x}+c=\frac{1}{4}ln|z^4+  1|

    Ae^{\frac{4}{x}}=|z^4+1|

    I get y=\sqrt[4]{x^4(Ae^{\frac{4}{x}}-1})
    you have to integrate 1/x as well
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by MrGreedy)
    you have to integrate 1/x as well
    Doh yes forgot about integating LHS!
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by MrGreedy)
    you have to integrate 1/x as well
    So I now get y=\sqrt[4]{x^4(Ax^4-1)}. How would I verify that this is correct?
    Offline

    9
    ReputationRep:
    (Original post by TheEd)
    So I now get y=\sqrt[4]{x^4(Ax^4-1)}. How would I verify that this is correct?
    substitute it back into the original differental equation
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by MrGreedy)
    substitute it back into the original differental equation
    Damn can't get it to verify! Must be wrong!
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by MrGreedy)
    substitute it back into the original differental equation
    When I'm down to this stage:

    |x|^4+4c=|\frac{y^4}{x^4}+1|

    how would I get rid of the mod signs?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 3, 2010
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
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.