User:Egm6321.f11.team4.allen/HW2

= Problem 2.14 - Solve the general L1-ODE-VC =

From [[media:Pea1.f11.mtg6.djvu|Mtg 11-5]]

From [[media:Pea1.f11.mtg6.djvu|Mtg 12-1]]

Given
The general structure of Linear 1st-Order ODEs with Varying Coefficients (L1-ODE-VC) is

$$\displaystyle a_1(x)y'+a_0(x)y=b(x). $$

Find
The solution for $$\displaystyle y $$ if:

$$\displaystyle 1) a_1(x)=1, a_0(x)=x, b(x)=2x+3$$.

$$\displaystyle 2) $$ In general for $$\displaystyle a_1(x), a_0(x), b(x)$$.

$$\displaystyle 3) a_1(x)=x^2+1, a_0(x)=x, b(x)=2x$$.

Solution
$$\displaystyle 1) $$ If $$\displaystyle a_1(x)=1, a_0(x)=x, b(x)=2x+3$$ then $$\displaystyle y'+xy=2x+3$$ then

The integrating factor is $$\displaystyle h= \exp \int^{x}{a_0(x)ds}$$ which is $$\displaystyle h= \exp \int^{x}{sds}$$.

So, the integrating factor is $$\displaystyle h=exp\frac{x^2}{2}$$.

Using the integrating factor and eq. (1) 11-5 yields

$$ y=\exp\frac{-x^2}{2}\int^{x}{\exp\frac{-x^2}{2}*(2s+3)}ds$$.

Solving for $$y$$ yields

$$ y=\exp\frac{-x^2}{2}[2\int^{x}\exp\frac{s^2}{2}sds+3\int^{x}\exp\frac{s^2}{2}ds]$$

$$ y=2x+\exp\frac{-x^2}{2}[3\int^{x}\exp\frac{s^2}{2}ds]+C$$

$$\displaystyle 2) $$ In general for $$\displaystyle a_1(x), a_0(x), b(x)$$.

The integrating factor is $$\displaystyle h=\exp[\int^{x}\frac{a_0(s)}{a_1(s)}ds]$$.

Using eq. (1) 11-5 and replacing the integrating factor and right side function with the general terms, yields

$$\displaystyle y(x)=\frac{1}{\exp[\int^{x}{\frac{a_0(s)}{a_1(s)}}ds]}*\int^{x}\exp[\int^{x}{\frac{a_0(s)}{a_1(s)}ds}]*\frac{b(s)}{a_1(s)}ds$$.

$$\displaystyle3)$$ If $$\displaystyle a_1(x)=x^2+1, a_0(x)=x, b(x)=2x$$ then

The integrating factor is

$$\displaystyle h=\exp[\int^{x}\frac{s}{s^2+1}ds]$$.

This simplifies to

$$\displaystyle h=\exp*\frac{1}{2}\ln(x^2+1)=(x^2+1)^{1/2}$$.

So, the L1-ODE-VC becomes

$$\displaystyle (x^2+1)^{1/2}y'+(x^2+1)^{1/2}*\frac{x}{x^2+1}y=(x^2+1)^{1/2}*\frac{2x}{x^2+1}$$.

Simplifying reduces it to

$$\displaystyle (x^2+1)^{1/2}\frac{dy}{dx}+\frac{x}{(x^2+1)^\frac{1}{2}}y=\frac{2x}{(x^2+1)^\frac{1}{2}}$$.

Rearranging yields

$$\displaystyle (x^2+1)^{1/2}dy=\frac{x(2-y)}{(x^2+1)^\frac{1}{2}}dx$$.

Differenting both sides yields

$$\displaystyle (x^2+1)^\frac{1}{2}y=(2-y)*(x^2+1)^\frac{1}{2}$$.

Multiply through by $$\displaystyle (x^2+1)^\frac{1}{2}$$ and rearrange yields

$$\displaystyle y=\frac{C}{(x^2+1)^\frac{1}{2}}$$.

= Problem 2.15 - Show That K1 In (3) p.11-4 Is NOT Necessary = From [[media:Pea1.f11.mtg12.djvu|Mtg 12-2]]

Given
Eq. (3) p.11-3 is

$$\displaystyle y'+a_0(x)y=b(x)$$.

Eq. (3) p.11-4 is

$$\displaystyle h(x)=\exp[\int^x{a_0(s)ds+k_1}]$$.

Find
Show that the integration constant $$\displaystyle k_1$$ in (3) p.11-4 is NOT necessary.

Solution
Multiply (3) p.11-3 by the integrating factor, (3) p.11-4, which yields the following:

$$\displaystyle \exp[\int^x{a_0(s)ds+k_1}]*y'+\exp[\int^x{a_0(s)ds+k_1}]*a_0(x)y=\exp[\int^x{a_0(s)ds+k_1}]*b(x)$$.

Next, seperate $$\displaystyle \exp[\int^x{a_0(s)ds+k_1}]$$ into $$\displaystyle \exp[\int^x{a_0(s)ds}]*\exp[k_1]$$, which yields the following:

$$\displaystyle \exp[k_1]*\exp[\int^x{a_0(s)ds}]*y'+\exp[k_1]*\exp[\int^x{a_0(s)ds}]*a_0(x)y=\exp[k_1]*\exp[\int^x{a_0(s)ds}]*b(x)$$.

$$\displaystyle \exp[k_1]$$ cancels out of the equation showing $$\displaystyle k_1$$ is NOT necessary.

Author
[Author]

= Problem 2.16 - Show Solution from the Lecture Agrees with King 2003 = From [[media:Pea1.f11.mtg12.djvu|Mtg 12-2]]

Given
Eq. (3) p.11-3 is

$$\displaystyle y'+a_0(x)y=b(x)$$.

Eq. (3) p.11-4 is

$$\displaystyle h(x)=\exp[\int^x{a_0(s)ds+k_1}]$$.

Eq. (1) p.11-5 is

$$\displaystyle y(x)=\frac{1}{h(x)}[\int^x[h(s)b(s)ds+k_2]$$.

Find
Show that the solution of (3) p.11-3 in (1) p.11-5 agrees with the result presented in King 2003 p.512.

$$\displaystyle y(x)=Ay_H(x)+y_P(x)$$

Solution
King defines the L1-ODE-VC as

$$\displaystyle y'+P(x)y=Q(x)$$.

King also defines the integrating factor as

$$\displaystyle h(x)=\exp[\int^x{P(t)dt}]$$ where t is a dummy variable.

The equation $$\displaystyle y(x)=Ay_H(x)+y_P(x)$$ is defined with

$$\displaystyle y_H(x)=A*\exp[-\int^x{P(t)}dt]$$

$$\displaystyle y_P(x)=\exp[-\int^x{P(t)dt}]\int^x{Q(s)}\exp[\int^x{P(t)dt}]ds$$

and $$\displaystyle A$$ is constant of integration.

Expand Equation (1) P.11-5 with Equations (3) p.11-3 and (3) p.11-4 which yields

$$\displaystyle y(x)=\frac{1}{\exp[\int^x{a_0(s)ds}]}*[\int^x{[\exp[\int^x{a_0(s)ds}]*b(s)ds]+k_2}]$$

Further expand the above equation and you can break up $$\displaystyle y_H(x)$$ as

$$\displaystyle y_H(x)=\frac{k_2}{\exp[\int^x{a_0(s)ds}]}$$.

This shows that $$\displaystyle k_2=A$$ and $$\displaystyle a_0(s)=P(t)$$.

Also, it can be shown that $$\displaystyle b(s)=Q(t)$$ from the equation below:

$$\displaystyle y_P(x)=\frac{1}{\exp[\int^x{a_0(s)ds}]}*[\int^x{[\exp[\int^x{a_0(s)ds}]*b(s)ds]}]$$.

Author
Allen