User:Eml5526.s11.team3.akj/Homework 2

=--Problem 2.4--=

Problem Statement
To solve for a G1DM1.0/D1 given in (3) in P 9-2

Given

An elastic bar of unit length is applied with a distributed load of $$f(x)=5x$$  along the axis of the bar, and linear variation of product of elastic modulus and area of the bar is given as  $$a_{2}(x)=2+3x$$

Boundary conditions

1) Essential boundary condition $$\gamma _{g}={1}, g=4$$ , which implies  $$ u(1)=4$$

2) Natural boundary condition $$\gamma _{h}={1}, h=6$$ , which implies  $$-\frac{\partial u(x=0)}{\partial x} =6 $$

Hence the above problem reduces to finding exact solution to the differential equation given by,

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}+5x=0 $$

$$ \forall x \in ] 0,1 [ $$

to solve the above differential equation we first solve the following integrals

1) $$\int lnxdx$$ 2) $$\int xlnxdx$$ 3) $$\int \frac{x^2}{1+cx}$$ 4) $$\int \frac{x^2}{a+bx}$$ 5) Find the exact solution of

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}+5x=0 $$

$$ \forall x \in ] 0,1 [ $$

$$ u(1)=4, -\frac{\partial u(x=0)}{\partial x} =6 $$

6) Plot u(x) 7) Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

Solution
1) To the solve the above integral we use, Integration by parts {| style="width:100%" border="0" $$ \int{u}dv =uv-\int{v}du $$ {| style="width:100%" border="0" where,
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$$\int lnxdx=xlnx-\int \frac{x}{x}dx$$

$$=xlnx-\int 1dx$$

2)

we again use Integration by parts

here,

$$\int xlnxdx=\frac{x^2}{2}lnx-\int \frac{x^2}{2x}dx$$

$$=\frac{x^2}{2}lnx-\int \frac{x}{2}dx$$

$$=\frac{x^2}{2}lnx- \frac{x^2}{4}$$

3) here,

hence,

$$\int \frac{x^2}{1+cx}= \frac{x^2}{c}ln(1+cx)-\int \frac{2}{c}ln(1+cx)xdx$$

for, $$\int ln(1+cx)xdx$$

hence,

$$= \frac{x^2}{c}ln(1+cx)-\frac{2}{c}[[\frac{x}{c}(1+cx)ln(1+cx)-\frac{x}{c}(1+cx)]-\int \frac{((1+cx)ln(1+cx)-(1+cx))}{c}dx]$$

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{2}{c^2}\int (1+cx)ln(1+cx)dx-\frac{2}{c^2}\int (1+cx)dx$$

for, $$\int (1+cx)ln(1+cx)dx$$

hence,

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{2}{c^2}[\frac{(1+cx)^2ln(1+cx)}{2c}-\int \frac{(1+cx)^2cdx}{2c(1+cx)}]-\frac{2}{c^2}\int (1+cx)dx$$

$$= \frac{x^2}{c}ln(1+cx)-\frac{2x}{c^2}(1+cx)ln(1+cx)+\frac{2x}{c^2}(1+cx)+\frac{(1+cx)^2}{c^3}ln(1+cx)-\frac{1}{c^2}\int (1+cx)dx-\frac{2}{c^2}\int (1+cx)dx$$

$$=ln(1+cx)[\frac{x^2}{c}-\frac{2x}{c^2}(1+cx)+\frac{(1+cx)^2}{c^3}]+\frac{2x}{c^2}(1+cx)-\frac{3}{c^2}\int (1+cx)dx$$

$$=ln(1+cx)[\frac{x^2}{c}-\frac{2x}{c^2}-\frac{2x^2c}{c^2}+\frac{1}{c^3}+\frac{c^2x^2}{c^3}+\frac{2cx}{c^3}]+\frac{2x}{c^2}(1+cx)-\frac{3}{c^2}[\frac{(1+cx)^2}{2c}]$$

$$=ln(1+cx)[\frac{1}{c^3}]+\frac{2x}{c^2}+\frac{2x^2}{c}-\frac{3}{2c^3}[1+c^2x^2+2cx]$$

$$=ln(1+cx)[\frac{1}{c^3}]+\frac{2x}{c^2}+\frac{2x^2}{c}-\frac{3}{2c^3}-\frac{3x^2}{2c}-\frac{3x}{c^2}$$

$$=\frac{1}{c^3}ln(1+cx)+\frac{x^2}{2c}-\frac{x}{c^2}-\frac{3}{2c^3}$$

$$=\frac{2ln(1+cx)+x^2c^2-2cx}{2c^3}+constant$$

4)

here,

hence,

$$\int \frac{x^2}{a+bx}= \frac{x^2}{b}ln(a+bx)-\int \frac{2}{b}ln(a+bx)xdx$$

for, $$\int ln(a+bx)xdx$$

hence,

$$= \frac{x^2}{b}ln(a+bx)-\frac{2}{b}[[\frac{x}{b}(a+bx)ln(a+bx)-\frac{x}{b}(a+bx)]-\int \frac{((a+bx)ln(a+bx)-(a+bx))}{b}dx]$$

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{2}{b^2}\int (a+bx)ln(a+bx)dx-\frac{2}{b^2}\int (a+bx)dx$$

for, $$\int (a+bx)ln(a+bx)dx$$

hence,

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{2}{b^2}[\frac{(a+bx)^2ln(a+bx)}{2b}-\int \frac{(a+bx)^2bdx}{2b(a+bx)}]-\frac{2}{b^2}\int (a+bx)dx$$

$$= \frac{x^2}{b}ln(a+bx)-\frac{2x}{b^2}(a+bx)ln(a+bx)+\frac{2x}{b^2}(a+bx)+\frac{(a+bx)^2}{b^3}ln(a+bx)-\frac{1}{b^2}\int (a+bx)dx-\frac{2}{b^2}\int (a+bx)dx$$

$$=ln(a+bx)[\frac{x^2}{b}-\frac{2x}{b^2}(a+bx)+\frac{(a+bx)^2}{b^3}]+\frac{2x}{b^2}(a+bx)-\frac{3}{b^2}\int (a+bx)dx$$

$$=ln(a+bx)[\frac{x^2}{b}-\frac{2ax}{b}-\frac{2x^2b}{b^2}+\frac{a^2}{b^3}+\frac{b^2x^2}{b^3}+\frac{2abx}{b^3}]+\frac{2x}{b^2}(a+bx)-\frac{3}{b^2}[\frac{(a+bx)^2}{2b}]$$

$$=ln(a+bx)[\frac{a^2}{b^3}]+\frac{2ax}{b^2}+\frac{2x^2}{b}-\frac{3}{2b^3}[a^2+b^2x^2+2abx]$$

$$=ln(a+bx)[\frac{a^2}{b^3}]+\frac{2ax}{b^2}+\frac{2x^2}{b}-\frac{3a^2}{2b^3}-\frac{3x^2}{2b}-\frac{3ax}{b^2}$$

$$=\frac{a^2}{b^3}ln(a+bx)+\frac{x^2}{2b}-\frac{ax}{b^2}-\frac{3a^2}{2b^3}$$

$$=\frac{2a^2ln(a+bx)+x^2b^2-2bax}{2b^3}+constant$$

5)

$$ \frac{\partial [(2+3x)\frac{\partial u}{\partial x}]}{\partial x}=-5x $$

Integrating the above expression we get,

$$ (2+3x)\frac{\partial u}{\partial x}=\frac{-5x^2}{2}+k1$$

$$\frac{\partial u}{\partial x}=\frac{-5x^2}{2(2+3x)}+\frac{k1}{2+3x}$$

given,

$$-\frac{\partial u(x=0)}{\partial x} =6$$

$$-6= 0+ \frac{k1}{2}$$

hence $$k1=-12$$

$$\frac{\partial u}{\partial x}=\frac{-5x^2}{2(2+3x)}+\frac{-12}{2+3x}$$

$$\partial u=\frac{-5x^2}{2(2+3x)}\partial x+\frac{k1}{2+3x}\partial x$$

Integrating the above equation

$$u=\frac{-5}{2}\int \frac{x^2}{2+3x}dx-12\int \frac{dx}{2+3x}$$

from (Eq.2) $$\int \frac{x^2}{a+bx}dx=\frac{2a^2ln(a+bx)+bx(bx-2a)}{2b^3}$$

hence $$\int \frac{x^2}{2+3x}dx=\frac{8ln(2+3x)+3x(3x-4)}{54}$$

$$u=\frac{-5}{2}[\frac{8ln(2+3x)+3x(3x-4)}{54}]-12[\frac{ln(2+3x)}{3}]+k2$$

$$u=\frac{-118}{27}ln(2+3x)-\frac{5x}{36}(3x-4)+k2$$

given $$ u(1)=4$$

$$4=\frac{-118}{27}ln(5)-\frac{5}{36}(-1)+k2$$

$$k2=10.89495$$

hence the exact solution is

6) Plot of u(x)

[[File:Fe1.s11.team3.HW2.fig2.jpg|frame|center|Fe1.s11.team3.HW2.fig2]

7)

The above integral is a special case if (Eq.2) where  $$a=b=1$$

hence,

$$\int \frac{x^2}{1+x}= \frac{2*1^2ln(1+x)+x(x-2)}{2*1^3}+constant$$

Problem Solved by akj.