Join TSR now and get all your revision questions answeredSign up now
    • Thread Starter
    Offline

    3
    ReputationRep:
    I've no idea where to start on question 6b and question 7. I don't think I even know what they're asking.
    Attached Files
  1. File Type: doc Doc1.doc (268.5 KB, 48 views)
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by maggiehodgson)
    I've no idea where to start on question 6b and question 7. I don't think I even know what they're asking.
    For 6(b) you are simply required to show that y=\text{e}^{k_1x} satisfies the d.e.
    so put y=\text{e}^{k_1x} find \frac{\text{d}y}{\text{d}x}\text  { and }\frac{\text{d}^2y}{\text{d}x^2} substirute into the left hand side of the d,e, and show that it equals 0.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by brianeverit)
    For 6(b) you are simply required to show that y=\text{e}^{k_1x} satisfies the d.e.
    so put y=\text{e}^{k_1x} find \frac{\text{d}y}{\text{d}x}\text  { and }\frac{\text{d}^2y}{\text{d}x^2} substirute into the left hand side of the d,e, and show that it equals 0.
    Thanks. I thought I would have had to start from the DE end and work towards the y=...

    This help might move me forward for question 7 now. Thank you.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by brianeverit)
    For 6(b) you are simply required to show that y=\text{e}^{k_1x} satisfies the d.e.
    so put y=\text{e}^{k_1x} find \frac{\text{d}y}{\text{d}x}\text  { and }\frac{\text{d}^2y}{\text{d}x^2} substirute into the left hand side of the d,e, and show that it equals 0.
    I'm afraid that question 6 has not helped with question 7. I'm using the AQA FP3 booklet from their website. On page 105 it goes into deriving the different form of solution. Would I have to do that or can you think of a simpler method?

    I'm attaching the start of the AQA booklet page to show you what I'm talking about.

    Thanks
    Attached Files
  2. File Type: doc Doc1.doc (268.5 KB, 29 views)
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by maggiehodgson)
    I'm afraid that question 6 has not helped with question 7. I'm using the AQA FP3 booklet from their website. On page 105 it goes into deriving the different form of solution. Would I have to do that or can you think of a simpler method?

    I'm attaching the start of the AQA booklet page to show you what I'm talking about.

    Thanks
    y=Ay_1+By_1\implies \frac{\text{d}y}{\text{d}x}=A \frac{\text{d}y_1}{\text{d}x}+B \frac{\text{d}y_2}{\text{d}x}
    Get the second derivative in the bsame way and sub into the d,e,. Then use the fact that you know  a\frac{\text{d}^2y_1}{\text{d}x^  2}+b\frac{\text{d}y_1}{\text{d}x  }+cy_1=0
    and  a\frac{\text{d}^2y_2}{\text{d}x^  2}+b\frac{\text{d}y_2}{\text{d}x  }+cy_2=0
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by brianeverit)
    y=Ay_1+By_1\implies \frac{\text{d}y}{\text{d}x}=A \frac{\text{d}y_1}{\text{d}x}+B \frac{\text{d}y_2}{\text{d}x}
    Get the second derivative in the bsame way and sub into the d,e,. Then use the fact that you know  a\frac{\text{d}^2y_1}{\text{d}x^  2}+b\frac{\text{d}y_1}{\text{d}x  }+cy_1=0
    and  a\frac{\text{d}^2y_2}{\text{d}x^  2}+b\frac{\text{d}y_2}{\text{d}x  }+cy_2=0

    Is this it?
    Attached Images
  3. File Type: pdf Scan0002.pdf (353.7 KB, 31 views)
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by maggiehodgson)
    Is this it?
    Yes. To complte it just take out the factors of A and B and observe that the contents of the brackets are zero since y1 and y2 are solutions of the d.e.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by brianeverit)
    Yes. To complte it just take out the factors of A and B and observe that the contents of the brackets are zero since y1 and y2 are solutions of the d.e.

    Many thanks. I think it's beginning to make sense but, for some reason, I feel that I should have to work towards y = Ay1 + By2 rather than starting from there and showing that it works.
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by maggiehodgson)
    Many thanks. I think it's beginning to make sense but, for some reason, I feel that I should have to work towards y = Ay1 + By2 rather than starting from there and showing that it works.
    Definitely not. The method I have suggested is the standard method.
    Google "Differentialm equations, linear combination of solutions" and look at some of the articles.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by brianeverit)
    Definitely not. The method I have suggested is the standard method.
    Google "Differentialm equations, linear combination of solutions" and look at some of the articles.
    Oh no, I believe you. It's me just thinking the wrong way.

    Thanks for all you help over this and several other questions in the past. It's very much appreciated.

    Happy New Year
    • Study Helper
    Offline

    3
    ReputationRep:
    (Original post by maggiehodgson)
    Oh no, I believe you. It's me just thinking the wrong way.

    Thanks for all you help over this and several other questions in the past. It's very much appreciated.

    Happy New Year
    Same to you and I'm just pleased to be able to help.
 
 
 
Poll
Which pet is the best?
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

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