University of Florida/Egm4313/s12.team11.perez.gp/R5.9

Consider the L2-ODE-CC (5) p.7b-7 with ${\displaystyle log(1+x)\!}$ as excitation:

${\displaystyle y''-3y'+2y=r(x)\!}$ (5) p.7b-7

${\displaystyle r(x)=log(1+x)\!}$ (1) p.7c-28

and the initial conditions

${\displaystyle y({\frac {-3}{4}})=1,y'({\frac {-3}{4}})=0\!}$.

Project the excitation ${\displaystyle r(x)\!}$ on the polynomial basis

${\displaystyle {b_{j}(x)=x^{j},j=0,1,...,n}\!}$ (1)

i.e., find ${\displaystyle d_{j}\!}$ such that

${\displaystyle r(x)\approx r_{n}(x)=\sum _{j=0}^{n}d_{j}x^{j}\!}$ (2)

for x in ${\displaystyle [{\frac {-3}{4}},3]\!}$, and for n = 3, 6, 9.

Plot ${\displaystyle r(x)\!}$ and ${\displaystyle r_{n}(x)\!}$ to show uniform approximation and convergence.

Note that:

${\displaystyle \left\langle x^{i},r\right\rangle =\int _{a}^{b}x^{i}log(1+x)dx\!}$ (3)

Using Matlab, this is the code that was used to produce the results:

Find ${\displaystyle y_{n}(x)\!}$ such that:

${\displaystyle y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\!}$ (1) p.7c-27

with the same initial conditions as in (2) p.7c-28.

Plot ${\displaystyle y_{n}(x)\!}$ for n = 3, 6, 9, for x in ${\displaystyle [{\frac {-3}{4}},3]\!}$.

In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Using integration by parts, and then with the help of of

General Binomial Theorem

${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}\!}$

For ${\displaystyle n=0\!}$:

${\displaystyle \int x^{0}log(1+x)dx=\int log(1+x)dx\!}$

For substitution by parts, ${\displaystyle u=log(1+x),du={\frac {1}{1+x}},dv=dx,v=x\!}$

${\displaystyle \int log(1+x)dx=xlog(1+x)-\int {\frac {x}{1+x}}dx\!}$

${\displaystyle \int log(1+x)dx=xlog(1+x)-\int (1-{\frac {1}{1+x}})dx\!}$

${\displaystyle \int log(1+x)dx=xlog(1+x)-x+log(1+x)+C\!}$

Therefore:

                                     ${\displaystyle \int log(1+x)dx=(x+1)log(1+x)-x+C\!}$

Using the General Binomial Theorem:

${\displaystyle (x+y)^{0}=\sum _{k=0}^{0}{\binom {0}{k}}x^{0-k}y^{k}=1\!}$

Therefore: ${\displaystyle \int (1)log(1+x)dx=\int log(1+x)dx\!}$

Which we have previously found that answer as:

                                     ${\displaystyle \int log(1+x)dx=(x+1)log(1+x)-x+C\!}$

For ${\displaystyle n=1\!}$:

${\displaystyle \int x^{1}log(1+x)dx=\int xlog(1+x)dx\!}$

Initially we use the following substitutions: ${\displaystyle t=1+x,x=t-1,dt=dx\!}$

${\displaystyle \int xlog(1+x)dx=\int (t-1)log(t)dt=\int (tlog(t)-\log(t))dt\!}$

First let us consider the first term: ${\displaystyle \int tlog(t)dt\!}$

Next, we use the integration by parts: ${\displaystyle u=\log {t},du={\frac {1}{t}}dt,dv=tdt,v={\frac {1}{2}}t^{2}\!}$
${\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}t^{2}({\frac {1}{t}}dt)\!}$

${\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-\int {\frac {1}{2}}tdt)\!}$

${\displaystyle \int tlog(t)dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}\!}$

Next let us consider the second term: ${\displaystyle \int log(t)dt\!}$

Again, we will use integration by parts: ${\displaystyle u=\log {t},du={\frac {1}{t}}dt,dv=dt,v=t\!}$
${\displaystyle \int tlog(t)dt=tlog(t)-\int t({\frac {1}{t}}dt)\!}$

${\displaystyle \int tlog(t)dt=tlog(t)-\int dt\!}$

${\displaystyle \int tlog(t)dt=tlog(t)-t\!}$

Therefore:

${\displaystyle \int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-(tlog(t)-t)\!}$

${\displaystyle \int (tlog(t)-\log(t))dt={\frac {1}{2}}t^{2}log(t)-{\frac {1}{4}}t^{2}-tlog(t)+t\!}$

Re-substituting for t:

${\displaystyle \int xlog(1+x)dx={\frac {1}{2}}(1+x)^{2}log(1+x)-{\frac {1}{4}}(1+x)^{2}-(1+x)log(1+x)+(1+x)+C\!}$

${\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}(1+x)log(1+x)-{\frac {1}{4}}(1+x)-log(1+x)+1)+C\!}$

${\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}$

Therefore:

                                     ${\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}$

Using the General Binomial Theorem for the integral with t substitution ${\displaystyle \int (t-1)log(t)dt\!}$:

${\displaystyle (x+y)^{1}=(t+(-1))^{1}=(t+(-1))=\sum _{k=0}^{1}{\binom {1}{k}}x^{1-k}y^{k}=\sum _{k=0}^{1}{\binom {1}{k}}t^{1-k}(-1)^{k}=t-1\!}$

Therefore: ${\displaystyle \int (t-1)log(t)dt=\int xlog(1+x)dx\!}$

Which we have previously found that answer as:

                                     ${\displaystyle \int xlog(1+x)dx=(1+x)({\frac {1}{2}}xlog(1+x)-{\frac {1}{2}}log(1+x)-{\frac {1}{4}}x+{\frac {3}{4}})+C\!}$