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: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 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 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:

$\displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0$

(6.5.5)

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

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

(6.5.6)

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

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

(6.5.7)

In order for 6.5.6 to be true it can be seen that

$\displaystyle r=2$

(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:

$\displaystyle u_1(x)=e^{2x}$

(6.5.9)

Using equation (4) from page 34-5 of the class notes Mtg 34 (b) the following expression for the 2nd homogeneous solution is obtained:

$\displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$

(6.5.10)

Equation 6.5.1 can be rearranged into the desired form shown:

$\displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0$

(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:

$\displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$

(6.5.12)

Evaluating the inner most integral yields:

$\displaystyle u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$

(6.5.13)

Finally evaluating the last integral and simplifying yields:

$\displaystyle u_2(x)=\frac{-1}{4}(2x+3)$

(6.5.14)

This is the 2nd homogeneous solution.