Consider this notice for the test on Tuesday 13 March at 6.45 p.m (just under three weeks away) (note that there is still a small chance that this tell will be held on Thursday 8 March at 8.15 p.m.).

Please find a sample test here

Note that the format will be the same as this.

- Homogeneous Second Order Linear
- Homogeneous Second Order Linear with boundary/initial conditions
- Non-Homogeneous Second Order Linear
- Non-Homogeneous Second Order Linear with boundary/initial conditions
- Second Order Separable with Step and Impulse Functions
- Second Order Separable with Step and Impulse Functions with boundary/initial conditions

Q. 5 and 6 will be covered Thursday night. Note that for questions 1-4 the roots will not be complex — although they could be fractions or surds (i.e. ye might need the formula).

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May 17, 2012 at 4:48 pm

Student 17Im in the process of revising the differential equations question. Just a quick query… say there on question3(b)(ii) of the 2011/12 paper. When you get to ie …To differentiate this, is this the chain rule or the product??? On the solution its worked out as I myself think its the chain rule, am I correct? How does that work again?

May 17, 2012 at 4:54 pm

J.P. McCarthyFirst off, differentiation is linear which means that we can differentiate a sum of terms term-by-term and pull out constants:

In this case we have

(*)

Now and are compositions of functions so do need the Chain Rule (see “Chain Rule by Rule”: http://irishjip.wordpress.com/2010/10/26/the-chain-rule/)

“differentiate the outside function with respect to the inside, and then multiply by the derivative of the inside”

.

Similarly,

.

Going back to (*):

as you said.

Regards,

J.P.