The Student Room Group

Easy Second Order DE question (FP2)

ImageUploadedByStudent Room1474400246.417102.jpg

Hey guys i'm having a bit of trouble knowing what to do for part a of question 2? (Finding the value of K for 2 marks) Cheers!

do you find dx/dt etc and plug it back in?
Posted from TSR Mobile
(edited 7 years ago)
Original post by iMacJack


Hey guys i'm having a bit of trouble knowing what to do for part a of question 2? (Finding the value of K for 2 marks) Cheers!

do you find dx/dt etc and plug it back in?
Posted from TSR Mobile


Use the differentials of xx with respect to tt and plug them into the equation, then solve for k.
Original post by iMacJack
ImageUploadedByStudent Room1474400246.417102.jpg

Hey guys i'm having a bit of trouble knowing what to do for part a of question 2? (Finding the value of K for 2 marks) Cheers!

do you find dx/dt etc and plug it back in?
Posted from TSR Mobile


Find the derivatives, substitute into the equation and compare coefficients.
you need to find dx/dt and also d2x/dt2 and put them into the LHS... it has to equal the RHS.
Reply 4
okay thanks guys I've done this and I come away with dx/dr = -tke^-t and d2x/dt2 = t^2ke^-t

I plug these in and I don't get anything seemably solvable?


Posted from TSR Mobile
Original post by iMacJack
okay thanks guys I've done this and I come away with dx/dr = -tke^-t and d2x/dt2 = t^2ke^-t

I plug these in and I don't get anything seemably solvable?


Posted from TSR Mobile


Er, you got them wrong. ddt(ke−t)=−ke−t≠−tke−t\frac{d}{dt}(ke^{-t})=-ke^{-t} \not= -tke^{-t}
Reply 6
Original post by RDKGames
Er, you got them wrong. ddt(ke−t)=−ke−t≠−tke−t\frac{d}{dt}(ke^{-t})=-ke^{-t} \not= -tke^{-t}


Back to the basics then!! how is that the case? (Apologies we have been thrown into FP2 without doing C3/4 so I'm having to pick up all the differentiation rules as I go)


Posted from TSR Mobile
Original post by iMacJack
Back to the basics then!! how is that the case? (Apologies we have been thrown into FP2 without doing C3/4 so I'm having to pick up all the differentiation rules as I go)


Posted from TSR Mobile


Chain rule. (and maybe product rule if you REALLY need it for the differential with kk)

y=ke−ty=ke^{-t}

Let u=−t⇒dudt=−1u=-t \Rightarrow \frac{du}{dt}=-1

⇒y=keu⟶dydu=keu\Rightarrow y=ke^u \longrightarrow \frac{dy}{du}=ke^u

dydt=dydu⋅dudt=keu⋅−1=−ke−t\frac{dy}{dt}=\frac{dy}{du} \cdot \frac{du}{dt} = ke^u \cdot -1 = -ke^{-t}
Original post by iMacJack
Back to the basics then!! how is that the case? (Apologies we have been thrown into FP2 without doing C3/4 so I'm having to pick up all the differentiation rules as I go)


Posted from TSR Mobile


Do diff and integration in C3-4 to the tea before doing FP2. Your gna mislearn stuff and damage your foundation in calculus otherwise.


Posted from TSR Mobile
Reply 9
Original post by physicsmaths
Do diff and integration in C3-4 to the tea before doing FP2. Your gna mislearn stuff and damage your foundation in calculus otherwise.


Posted from TSR Mobile


agreed
Reply 10
Original post by physicsmaths
Do diff and integration in C3-4 to the tea before doing FP2. Your gna mislearn stuff and damage your foundation in calculus otherwise.


Posted from TSR Mobile


I don't have a choice - our school is making us do it this way! only other way would be to learn the calculus off my own back which I'd need to allocate time for which I don't really have right now


Posted from TSR Mobile
Reply 11
Can someone do this and tell me what they get by the way please? I get 5/3 and I don't have the solution! (value of K)


Posted from TSR Mobile
Original post by iMacJack
Can someone do this and tell me what they get by the way please? I get 5/3 and I don't have the solution! (value of K)


Posted from TSR Mobile


Yes, that's right.
Reply 13
Original post by NotNotBatman
Yes, that's right.


Awesome thank you, also I might be wrong but for the second part I got A = -5/12 and B = -1/4, then plugged that back into the comp function and particular integral to get x = -5/12 e^7t - 1/4 e^-3t + 5/3 e^-t


Posted from TSR Mobile
Original post by iMacJack
I don't have a choice - our school is making us do it this way! only other way would be to learn the calculus off my own back which I'd need to allocate time for which I don't really have right now


Posted from TSR Mobile


At the very least you NEED to learn the chain rule, product rule, quotient rule, integration by substitution, and integration by parts. Take a look at pages 24-29 (or pages 22-27 if you look at the numbers written at the bottom of the pages) of http://www.physicsandmathstutor.com/download/Maths/A-level/C3/Worksheets-Notes/Edexcel%20C3%20Notes.pdf, and at pages 12-21 (or 10-19 if using the numbers at the bottom of the pages) of http://www.physicsandmathstutor.com/download/Maths/A-level/C4/Worksheets-Notes/Edexcel%20C4%20Notes.pdf
Reply 15
Original post by HapaxOromenon3
At the very least you NEED to learn the chain rule, product rule, quotient rule, integration by substitution, and integration by parts. Take a look at pages 24-29 (or pages 22-27 if you look at the numbers written at the bottom of the pages) of http://www.physicsandmathstutor.com/download/Maths/A-level/C3/Worksheets-Notes/Edexcel%20C3%20Notes.pdf, and at pages 12-21 (or 10-19 if using the numbers at the bottom of the pages) of http://www.physicsandmathstutor.com/download/Maths/A-level/C4/Worksheets-Notes/Edexcel%20C4%20Notes.pdf


Awesome I'll get working on that, cheers


Posted from TSR Mobile
Original post by iMacJack
Awesome thank you, also I might be wrong but for the second part I got A = -5/12 and B = -1/4, then plugged that back into the comp function and particular integral to get x = -5/12 e^7t - 1/4 e^-3t + 5/3 e^-t


Posted from TSR Mobile


Incorrect, I think. Your CF should be Ae−4t+Be−3tAe^{-4t}+Be^{-3t} so I do not know where you pulled an e7te^{7t} from?
Original post by iMacJack
Awesome thank you, also I might be wrong but for the second part I got A = -5/12 and B = -1/4, then plugged that back into the comp function and particular integral to get x = -5/12 e^7t - 1/4 e^-3t + 5/3 e^-t


Posted from TSR Mobile



That's not what I got, the power of 7t seems wrong.

Original post by iMacJack
Awesome I'll get working on that, cheers


Posted from TSR Mobile


If you haven't done all of the differentiation and integration chapters, don't start on second order differential equations yet, solidify your knowledge on the necessary core stuff first.
(edited 7 years ago)
Reply 18
Original post by NotNotBatman
That's not what I got, the power of 7t seems wrong.



If you haven't done all of the differentiation and integration chapters, then don't start on second order differential equations yet, solidify your knowledge on the necessary core stuff first.


Yeah I didn't mean 7t! & we have gone over all the chain rule, product rule etc quickly, I get it, I just need to put it into practice a few more times get fully comfortable with it! Thanks, I was able to do the rest of the worksheet relatively easily! :smile:

Ill reattempt the question in a bit :smile:


Posted from TSR Mobile
(edited 7 years ago)
Original post by iMacJack
Awesome I'll get working on that, cheers


Posted from TSR Mobile


For second order DEs, you need to know the differentiation from C3/4 but not the integration. You'll need that for first oder ones though.

Quick Reply