User:Egm6321.f11.team1/report6.bensplaypen.R*6.5

= Problem R*6.4 For a L2-ODE-VC using variation of parameters find the 2nd homogeneous solution and compare =

Reference Lecture 35-4 found here:[[media:pea1.f11.mtg35.djvu|Mtg 35]]

Given
The following liner, second order, ordinary differential equation, with varying coefficients (L2-ODE-VC)

$$\displaystyle (x - 1) y'' - x y' + y = x$$ $$\displaystyle (6.4.1) $$

Find
A valid homogeneous solution and call it $$\displaystyle u_1 $$

Find the particular solution using variation of parameters method and compare it to $$\displaystyle e^{xr_2(x)} $$

Solution
Lets start by validating that the homogeneous solution selected below

$$\displaystyle u_1 = e^x $$ $$\displaystyle (6.4.2) $$

Use the trial solution:

$$ \displaystyle u_1(x) = e^{rx}$$ $$\displaystyle (6.4.3) $$

Differentiate the above to get $$ \displaystyle u'_1(x) = r e^{rx}$$ $$\displaystyle (6.4.4)$$

$$ \displaystyle u''_1(x) = r^2 e^{rx} $$ $$\displaystyle (6.4.5)$$

Plugging these derivatives and the trial solution into 6.4.1 yields $$ \displaystyle \left[(x-1) r^2 - rx + 1\right] e^{rx} = 0$$ $$\displaystyle (6.4.6)$$

Rearranging by grouping the terms with x from above yields

$$ \displaystyle \left[(r^2 - r) x + (1 - r^2)\right] e^{rx} = 0$$ $$\displaystyle (6.4.7)$$

In order for the above equation to be equal to zero, the following equations must be satisfied

$$ \displaystyle r^2 - r = r (r - 1) = 0 $$

$$ \displaystyle 1 - r^2 = -(r + 1)(r - 1) = 0 $$ $$\displaystyle (6.4.8)$$

The solution of the above equations is $$\displaystyle r=1$$. So 6.4.3 can be written as

$$ \displaystyle u_1(x) = e^x $$ $$\displaystyle (6.4.9)$$

Which is a valid homogeneous solution

So now lets find the 2nd homogeneous solution using variation of parameters

We can rewrite 6.4.1 as

$$ \displaystyle y'' + \underbrace{\frac{-x}{x-1}}_{a_1(x)} y' + \underbrace{\frac{1}{x-1}}_{a_0(x)} y  = \underbrace{\frac{x}{x-1}}_{f(x)} $$ $$\displaystyle (6.4.10) $$

Using equation (4) from page 34-5 of the class notes [[media:pea1.f11.mtg34.djvu|Mtg 34 (b)]] the following expression for the 2nd homogeneous solution is obtained:

$$\displaystyle {u}_{2}(x)= \underbrace{{u}_{1}(x)}_{{e}^{x}} \int \frac{1}{ u_1^2(x)}exp[-\int \underbrace{a_1(x)}_{\frac{-(x)}{x-1}} \ dx] \ dx $$ $$\displaystyle (6.4.11) $$

Using Wolfram Alpha to evaluate both integrals and simplify it can be shown that

$$\displaystyle {u}_{2}(x)= x $$ $$\displaystyle (6.4.12)$$

Which is the 2nd homogeneous solution

This can can then be compared to $$\displaystyle e^{xr_2(x)} $$ as requested

if $$\displaystyle r_2(x) = ln(x)/x $$ the expressions for $$\displaystyle{u}_{2}$$ are equivalent

= R*6.5: Find Two Homogeneous Solutions using a Trial Solution =

Reference lecture notes page 36-3 found here [[media:pea1.f11.mtg36.djvu|Mtg 36]]

Given
Find u1 and u2 of Equation 6.5.1 below using the trial solution given in equation 6.5.2:

$$\displaystyle (x +1) y'' - (2x+3)y' + 2y = 0 $$ (6.5.1)

$$\displaystyle y(x)=e^{rx}$$ (6.5.2)

Solution
Lets start by differentiating the trial solution as follows:

$$ \displaystyle y'(x) = r e^{rx}$$ (6.5.3)

$$ \displaystyle y''(x) = r^2 e^{rx}$$ (6.5.4)

The next step is to plug the derivatives and the original trail function into equation 6.5.1 as follows:

$$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $$ <p style="text-align:right">(6.5.5)

Divide $$ \displaystyle e^{rx} $$ by both side and group all the $$ \displaystyle x $$ terms to obtain:

$$ \displaystyle x(r^2-2r)+(r-2)(r-1)=0$$ <p style="text-align:right">(6.5.6)

The following conditions must be satisfy for equation 6.5.6 to be true:

$$ \displaystyle r(r-2)=(r-2)(r-1)=0 $$ <p style="text-align:right"> (6.5.7)

In order for 6.5.6 to be true it can be seen that $$ \displaystyle r=2 $$ <p style="text-align:right">(6.5.8)

The first homogeneous solution is obtained by plugging equation 6.5.8 into the trial solution given in equation 6.5.2:

$$ \displaystyle u_1(x)=e^{2x} $$ <p style="text-align:right">(6.5.9)

Using equation (4) from page 34-5 of the class notes [[media:pea1.f11.mtg34.djvu|Mtg 34 (b)]] the following expression for the 2nd homogeneous solution is obtained:

$$ \displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$$ <p style="text-align:right">(6.5.10)

Equation 6.5.1 can be rearranged into the desired form shown:

$$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0$$ <p style="text-align:right"> (6.5.11)

The value for $$ \displaystyle a_1 $$ from equation 6.5.11 is then substituted into equation 6.5.10, additionally the first homogeneous solution from equation 6.5.9 is substituted to solve for the second homogeneous solution as shown:

$$ \displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$$ <p style="text-align:right"> (6.5.12)

Evaluating the inner most integral yields:

$$ \displaystyle u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$$ <p style="text-align:right">(6.5.13)

Finally evaluating the last integral and simplifying yields:

$$ \displaystyle u_2(x)=\frac{-1}{4}(2x+3)$$ <p style="text-align:right">(6.5.14)

This is the 2nd homogeneous solution.