FP2 Second Order Differential Equations - Practice Paper B

Hi there,

I am a gap year student so I have no teacher to help!

Part 7(C): Find the general solution of the differential equation in terms of V and X
Part 7(D): Write down the general solution in terms of Y and X

Given that y=vx
and that the answer to part C is:
V=Asin3x + Bcos3x + (X^2)/9 - (2/81)
Why is the answer to part D not:
Y=Bxcos3x + Axsin3x +(x^3)/9 -2x/81

??

Instead the answer is just:
Y=Bxcos3x + Axsin3x +(x^2)/9 -2/81 where x has not been multiplied onto the particular integrals?

Any help much appreciated!

Hi there,

I am a gap year student so I have no teacher to help!

Part 7(C): Find the general solution of the differential equation in terms of V and X
Part 7(D): Write down the general solution in terms of Y and X

Given that y=vx
and that the answer to part C is:
V=Asin3x + Bcos3x + (X^2)/9 - (2/81)
Why is the answer to part D not:
Y=Bxcos3x + Axsin3x +(x^3)/9 -2x/81

??

Instead the answer is just:
Y=Bxcos3x + Axsin3x +(x^2)/9 -2/81 where x has not been multiplied onto the particular integrals?

Any help much appreciated!

I'd be happy to help but could you scan or photograph the question?

We really need to see the question to help.

Your wish, is my command!

Your wish, is my command!

Edit: I've done the question too. My final general solution for $y$ in terms of $x$ agrees with yours $\vec{\mathcal{O}\mathcal{P}}$

$y(x) = \dfrac{x^3}{9} - \dfrac{2x}{81} + \mathcal{P}x\cos (3x) + \mathcal{Q}x\sin (3x)$ where $\mathcal{P}$ and $\mathcal{Q}$ are constants to be found.

Thanks.

Thank

I agree with the OP

(and checked by putting y = (x^3/9) - (2x/81) into the original DE to get x^5 back!)

Thank you for your help!



So do you both seem to think that the mark scheme is wrong?

I can't think why they would only multiply the x through the C.F and not the P.I !?