edexcel FP2

Iconic_panda

for this question, why is P.I t cube instead of t square, is there a way of knowing the power of t straight away?

(edited 6 years ago)

Reply 1

RDKGames

Study Forum Helper

Original post by Iconic_panda

for this question, why is P.I t cube instead of t square, is there a way of knowing the power of t straight away?

What's the question?

Reply 2

tiny hobbit

Original post by Iconic_panda

for this question, why is P.I t cube instead of t square, is there a way of knowing the power of t straight away?

I'll guess that you already have a t squared term in your CF. You therefore have to multiply by another t.

Similarly, if you already had a e^t term in the CF (just that, not multiplied by some more algebra), then you'd have your PI as te^t.

Reply 3

Iconic_panda

this question here

Original post by RDKGames

What's the question?

Screen Shot 2018-03-10 at 19.20.00.png29.3KB

Reply 4

Iconic_panda

my CF only have t in there, no t square, can you please check the question below?

Original post by tiny hobbit

Reply 5

RDKGames

Study Forum Helper

Original post by Iconic_panda

this question here

The auxiliary equation has a repeated root, 2, which is also in

3te^{2t}

The general rule of thumb, is that if your auxiliary eq. has a repeated root

\gamma

while your RHS of the ODE has

e^{\gamma t}

, then the suggested P.I. is

t^2e^{\gamma t}

. Here, we have the RHS as

te^{\gamma t}

therefore we go up one in the power of

t

on our P.I. as well, hence we use

t^3e^{\gamma t}

Have a good look here starting 'In a nutshell...' (particularly the P.I. choice section) to get a solid understanding of how to deal with all types of ODE's at A-Level.

(edited 6 years ago)

Reply 6

NotNotBatman

Original post by Iconic_panda

this question here

For the particular integral.

TRY

x= ate^{2t}

find

x'(t) = 2ate^{2t} + ae^{2t}

x''(t) = 4ate^{2t} + 4ae^{2t}

sub in ODE to produce a vanishing LHS : 0=0

oh no!

so we try

x=3t^2e^{2t}

Reply 7

Iconic_panda

ah right!
what if it keeps on not working hahha, do we keep going up the power?

Original post by NotNotBatman

For the particular integral.

TRY

x= ate^{2t}

find

x'(t) = 2ate^{2t} + ae^{2t}

x''(t) = 4ate^{2t} + 4ae^{2t}

sub in ODE to produce a vanishing LHS : 0=0

oh no!

so we try

x=3t^2e^{2t}

Reply 8

Iconic_panda

I got another question, what if the RHS is a combination of two, let say 5 sinx + e^x. would we try PI as a sin x+ b cos x +c e^x, i.e the combination of two as well?

Original post by RDKGames

The auxiliary equation has a repeated root, 2, which is also in

3te^{2t}

The general rule of thumb, is that if your auxiliary eq. has a repeated root

\gamma

while your RHS of the ODE has

e^{\gamma t}

, then the suggested P.I. is

t^2e^{\gamma t}

. Here, we have the RHS as

te^{\gamma t}

therefore we go up one in the power of

t

on our P.I. as well, hence we use

t^3e^{\gamma t}

Have a good look here starting 'In a nutshell...' (particularly the P.I. choice section) to get a solid understanding of how to deal with all types of ODE's at A-Level.

Reply 9

RDKGames

Study Forum Helper

Original post by Iconic_panda

I got another question, what if the RHS is a combination of two, let say 5 sinx + e^x. would we try PI as a sin x+ b cos x +c e^x, i.e the combination of two as well?