User:EGM6321.f09.Team3.cgs/HW3

Problem 1. Find m and n such that Equation 1 p 13-1 is exact

Equation 1 p13-1: $$x^{m}y^{n}[\sqrt{x}y''+2xy'+3y]$$

Step 1: Does Eq 1 satisfy Condition 1 for exactness?

Rewrite Eq 1 as follows:

Let p=y’

$$\underbrace{x^{(m+\frac{1}{2})}y^{n}}_{f(x,y,p)}y''+\underbrace{2x^{(m+1)}y^{n}p+3x^{m}y^{(n+1)}}_{g(x,y,p)}$$

$$ f(x,y,p)= x^{(m+\frac{1}{2})}y^{n}$$

$$ g(x,y,p)= 2x^{(m+1)}y^{n}p+3x^{m}y{(n+1)} $$

Eq 1 can be written as f(x,y,p)y”+ g(x,y,p)=0 therefore Condition 1 for exactness has been satisfied.

Step 2: Does Eq 1 satisfy Condition 2 for exactness?

Condition 2 for exactness requires the following:

$$f_{xx}+2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y} $$

and

$$f_{xp}+ pf_{yp} + 2f_{y} =g_{pp}$$ (Equation 3)

Evaluating each derivative, we obtain:

$$f_{x} = (m+\frac{1}{2})x^{(m-\frac{1}{2})}y^n $$

$$ f_{xx} = (m^2-\frac{1}{4})x^{(m-\frac{3}{2})}y^n $$

$$ f_{xy} = (m+\frac{1}{2})nx^{(m-\frac{1}{2})}y^{n-1} $$

$$ f_{y} =n x^{(m+\frac{1}{2})}y^{n-1} $$

$$ f_{yy} =n(n-1)x^{(m+\frac{1}{2})}y^{n-2} $$

$$ f_{xp} =f_{yp} =0 $$

$$ g_{x}= 2(m+1)x^{m}y^np+3mx^{(m-1)}y^{(n+1)} $$

$$ g_{xp}= 2(m+1)x^{m}y^n $$

$$ g_{y}= 2nx^{(m+1)}y^{(n-1)}p+3(n+1)x^{m}y^{n} $$

$$ g_{yp}= 2nx^{(m+1)}y^{(n-1)} $$

$$ g_{p}= 2x^{(m+1)}y^n $$

$$ g_{pp}= 0 $$

Substituting into Equation 3, we obtain:

$$ f_{xx} +2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y} $$

$$ (m^2-\frac{1}{4})x^{(m-\frac{3}{2})}y^n+2p(m+\frac{1}{2})nx^{(m-\frac{1}{2})}y^{(n-1)}+p^{2} n(n-1) x^{(m+\frac{1}{2})}y^{(n-2)}=2(m+1)x^{m}y^{n}+p^{2}nx^{(m+1)}y^{(n-1)}- 2nx^{(m+1)}y^{(n-1)}p+3(n+1)x^{m}y^{n}$$\\(Equation 4)

Substituting into Equation 4, we obtain:

$$ f_{xp} + pf_{yp} + 2f_{y} =g_{pp} $$

$$0+p(0)+2 n x^{(m+\frac{1}{2})}y^{(n-1)}=0 $$

In order for Equation 5 to equal 0, n must equal 0 Substitute n= 0 into equation 4 to determine m:

$$ (m^{2}-\frac{1}{4})x^{(m-\frac{3}{2})y^n}+2p(m+\frac{1}{2})(0)x^{(m-\frac{1}{2})}y^{(0-1)}+p^{2}(0)(0-1)x^{(m+\frac{1}{2})}y^{(0-2)}=2(m+1)x^{(m-1)}y^{0}+p^{2}(0)x^{(m+1)}y^{(0-1)}- 2(0)x^{(m+1)}y^{(0-1)}p+3(0+1)x^{m}y^{0} $$

$$m=\frac{1}{2} $$

Therefore the integrating factor is

$$h(x,y)= x^{(\frac{1}{2})}y^{0}=x^{(\frac{1}{2})} $$

and the equation that satisfies both exactness conditions is:

$$\sqrt{x}[\sqrt{x}y''+2xy'+3y] $$

or

$$x''+2\sqrt{x}xy'+3\sqrt{x}y $$

Problem 12

Problem: Attempt to develop the reduction of order method 2 using different algebraic operations:

1) $$ y(x)=U(x)\text{±}u_{1}(x)$$

2) $$ y(x)=U(x)/u_{1}(x)$$

3) $$ y(x)=u_{1}(x)/U(x)$$

$$ y''+a_{1}y'+a_{0}y=0$$   (Equation 1)

and a known variable $$ u_{1}(x)$$

Problem 12 Part 2

Given the following homogenous equation:

For algebraic operation 1, we will assume that

$$ y(x)=U(x)\text{±}u_{1}(x) $$ (Equation 2)

$$ y'(x)=U'(x)\text{±}u'_{1}(x) $$

$$ y(x)=U(x)\text{±}u''_{1}(x) $$

Substituting these values into Equation 1 has the following result:

$$ [U(x)\text{±}u_{1}(x)]+a_{1}[U'(x)\text{±}u'_{1}(x)]+a_{0}[U(x)\text{±}u_{1}(x)]=0$$

Regrouping the terms by order of U we find that the equation can now be written as:

$$ U(x)+a_{1}U'(x)+a_{0}U(x)\pm\underbrace{[u_{1}(x)+a_{1}u'_{1}(x)+u_{1}(x) }_{g(x,u,u',u'')} =0]$$ (Equation 3)

As $$u_{1}(x)$$ has been defined as an homogeneous function, then $$ g(x,u,u',u'')$$ is also homogeneous and can be set to 0 thus reducing Equation 3 to:

$$ U''(x)+a_{1}U'(x)+a_{0}U(x)$$

There has been no reduction of order, so this algebraic operation is not appropriate to solve the reduction of order problem.

Problem 12 Part 2

Given the algebraic operation

$$ \frac{U}{u_{1}}$$

Determine whether the algebraic operation can be used to reduce the order of Equation 1

$$y= \frac{U}{u_{1}} $$

$$y'= \frac{U'u_{1}-Uu_{1}'}{u_{1}^{2}} $$

Which can be reduced to

$$y'= \frac{U'}{u_{1}}-\frac{Uu_{1}'}{u_{1}^{2}} $$

Let $$g=Uu_{1}' $$

Let $$g'=U'u_{1}' +Uu_{1}''$$

$$y= \frac{U u_{1}- U' u_{1}'}{ u_{1}^{2}}-\frac{g'u_{1}^{2}- g(u_{1}^{2})'}{u_{1}^{4}}$$

$$y= \frac{U}{ u_{1}}- \frac{U' u_{1}'}{ u_{1}^{2}}-\frac{(U'u_{1}'+Uu_{1}'')u_{1}^{2}-Uu_{1}'(u_{1}^{2})'}{u_{1}^{4}}$$

Substituting into Equation 1:

$$ y''+a_{1}(y'+a_{0}y=0$$

$$\frac{U}{ u_{1}}- \frac{U' u_{1}'}{ u_{1}^{2}}-\frac{(U'u_{1}'+Uu_{1})u_{1}^{2}-Uu_{1}'(u_{1}^{2})'}{u_{1}^{4}} +a_{1}\frac{U'}{u_{1}}-\frac{Uu_{1}'}{u_{1}^{2}} )+a_{0}(\frac{U}{u_{1}} )=0$$

Assuming $$u_{1}(x)\neq 0$$ divide equation through by $$u_{1}^{4}$$

$$U u_{1}^{3}-2U' u_{1}' u_{1}^{2}+Uu_{1}u_{1}^{2}-U(u_{1}^{2})'u_{1}' +a_{1}(U' u_{1}^{2}-Uu_{1}^{2}u_{1}')+a_{0}Uu_{1}^{3} =0$$

Regrouping to separate out U, U’ and U’’ results in the following equation:

$$ U u_{1}^{3}+U'(-2u_{1}' u_{1}^{2} +a_{1}u_{1}^{2})+U(u_{1}u_{1}^{2}-(u_{1}^{2})'u_{1}'+ a_{0}u_{1}^{3}$$

Which can’t be reduced.

Problem 12 Part 3

Given the algebraic operation

$$ \frac{u_{1}}{U}$$

Determine whether the algebraic operation can be used to reduce the order of Equation 1

$$y= \frac{U_{1}}{U} $$

$$y'= \frac{u_{1}'U-u_{1}U'}{U^{2}} $$

Which can be reduced to

$$y'= \frac{u_{1}'}{U}-\frac{u_{1}U'}{U^{2}} $$

Let $$g=u_{1}U' $$

Let $$g'=u_{1}'U' u_{1}U''$$

$$y= \frac{u_{1} U- u_{1})' U'}{ U^{2}}-\frac{g'U^{2}- g(U^{2})'}{U^{4}}$$

$$y= \frac{u_{1}}{ U}- \frac{u_{1}' U'}{ U^{2}}-\frac{(u_{1}'U'+u_{1}U'')U^{2}-u_{1}U'(U^{2})'}{U^{4}}$$

Substituting into Equation 1:

$$ y''+a_{1}(y'+a_{0}y=0$$

$$\frac{u_{1}}{ U}- \frac{u_{1}' U'}{ U^{2}}-\frac{(u_{1}'U'+u_{1}U)U^{2}-u_{1}U'(U^{2})'}{U^{4}} +a_{1}\frac{u_{1}'}{U}-\frac{u_{1}U'}{U^{2}} )+a_{0}(\frac{u_{1}}{U} )=0$$

Problem ?? Assuming $$U(x)\neq$$ divide equation through by $$U^{4}$$

$$u_{1} U^{3}-2u_{1}' U' U^{2}+u_{1}UU^{2} -u_{1}(U^{2})'U'+a_{1}(u_{1}' U^{2}-u_{1}U^{2}U')+a_{0}u_{1}U^{3} =0$$

As this equation is non-linear, no reduction of order is possible.

Given

$$ (1-x^{2})y''-2xy'+2y=0 $$ (Equation 1)

Find solution using the trial solution

$$ y=ax^{b} $$

$$ y'=bax^{b-1} $$

$$ y''=b(b-1)ax^{b-2} $$

Substituting into Equation 1

$$ (1-x^{2}) ax^{b-2} -2x bax^{b-1} +2 ax^{b}  =0 $$ $$ (ax^{b-2}) -ax^{b} -2 abx^{b} +2 ax^{b}  =0 $$ $$ (ax^{b-2}) -ax^{b} -2 abx^{b} =0 $$ Assuming $$ ax^{b-2}/ne{0}$$, divide through by $$  ax^{b-2}$$ This results in the following equation:

$$ (x^{-2}) -1-2b =0 $$

Given

$$ (1-x^{2})y''-2xy'+2y=0 $$ (Equation 1) Find solution using the trial solution $$ y=er^{x} $$

$$ y'= re^{rx} $$

$$ y''= r^{2}e^{rx} $$

Substituting into Equation 1

$$ (1-x^{2}) r^{2}e^{rx} -2x re^{rx}  +2 e^{rx}  =0  $$ $$ r^{2}e^{rx}  - r^{2} x^{2} e^{rx}  -2rx e^{rx}  +2 e^{rx}=0 $$ Assuming $$ y=er^{x}/ne{0}$$, divide through by $$  er^{x}$$ This results in the following equation:

$$ r^{2} - r^{2} x^{2}-2rx +2 =0 $$