Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 15 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii

i

Solution

Differentiate

v^{'} = u

once, then substitute the other equation. The required second-order linear differential equation is

v^{''} = u^{'} = 4 v + 2, v^{''} - 4 v = 2.

ii

Solution

The characteristic equation of the homogeneous equation is

λ^{2} - 4 = 0

, giving exponential solutions

e^{2 t}

and

e^{- 2 t}

. A constant particular solution is

v_{p} = - 1/2

. Thus the general solution is

v (t) = A e^{2 t} + B e^{- 2 t} - \frac{1}{2}, A, B \in R .

iii

Solution

u = v^{'}

fixes

u

from the same two constants:

u (t) = 2 A e^{2 t} - 2 B e^{- 2 t} .

Indeed

u^{'} = 4 A e^{2 t} + 4 B e^{- 2 t} = 4 v + 2

, so both original equations hold. No further independent integration constant is available in this two-dimensional linear system of ordinary differential equations.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version