SEARCH THIS BLOG :-)

Sunday, 10 December 2006

CAMBRIDGE 3U YEAR 12 - EX 3F - Q6, 8, 10b, 14, 22, 23, - Simple Harmonic Motion - differential equation - selected solutions

QUESTIONS DONE:  Q6, 8, 10b, 14, 22, 23, 
  Q6 =========================
 
Q8 ==============

Q10b =========================
 
Q14 =========================
 
Q22 =========================
 
Q23 =========================
 

Sunday, 2 July 2006

SOLUTIONS TO 1997 COLLEGE ENTRANCE EXAM - mostly Calculus, Differentiation, Integration, Areas, Volumes.

SOLUTIONS TO 1997 COLLEGE ENTRANCE EXAM - mostly Calculus, Differentiation, Integration, Areas, Volumes.

Here are the questions. Then follow my solutions :-)













SOLUTIONS ===================

















Thursday, 1 June 2006

1. Consider the partial differential equation: $\displaystyle q(x) { \partial u\over \partial t}={ \partial \over \partial x} \left( p(x) { \partial u\over \partial x} \right)$ (a) Confirm that $$U(x) = A +(B-A){\int_0^x p(s)^{-1} ds\over \int_0^L p(s)^{-1} ds }$$ Is a solution of the partial differential equation (b) Confirm that if $u=U+v$ then $v$ indeed satisfies the partial differential equation if $u$ does (c) Confirm that $X(x)T(t)$ is indeed a solution of the partial diff eq if $X$ and $T$ are solutions of the following respectively: $${d\over dx} \left(p(x) {dX\over dx}\right) – kq(x)X = 0$$ $${dT\over dt} = kT$$ where $k$ is a constant

1.  Consider the partial differential equation:

$\displaystyle q(x) { \partial u\over \partial t}={ \partial \over \partial x} \left( p(x) { \partial u\over \partial x} \right)$
(a)   Confirm that
$$U(x) = A +(B-A){\int_0^x p(s)^{-1} ds\over \int_0^L p(s)^{-1} ds }$$
Is a solution of the partial differential equation

(b)   Confirm that if $u=U+v$ then $v$ indeed satisfies the partial differential equation if $u$ does

(c)   Confirm that $X(x)T(t)$ is indeed a solution of the partial diff eq if $X$ and $T$ are solutions of the following respectively:
$${d\over dx} \left(p(x) {dX\over dx}\right) – kq(x)X = 0$$
$${dT\over dt} = kT$$

where $k$ is a constant

SOLUTION ===============





Wednesday, 1 March 2006

QUESTION. Let $L:V\rightarrow W$ be a linear map. Let $w$ be an element of $W$. Let $\nu_0$ be an element of $V$ such that $L(\nu_0)=w$. Show that any solution of the equation $L(X)=w$ is a type $\nu_0+u$. where $u$ is an element of the kernel of $D$.


QUESTION. Let $L:V\rightarrow W$ be a linear map. Let $w$ be an element of $W$. Let $\nu_0$ be an element of $V$ such that $L(\nu_0)=w$. Show that any solution of the equation $L(X)=w$ is a type $\nu_0+u$. where $u$ is an element of the kernel of $D$.