User:Egm6321.f12.team1.pb/report6

=Problem 6.4* =

Given:
The characteristic equation

with $$\lambda = 5$$

and Euler L2-ODE-VC

Find:
1. $$a_2, a_1, a_0$$ such that ($$) is the characteristics equation of ($$)

2. 1st homogeneous solution $$y_1(x) = x^{\lambda}$$

3. Complete solution: $$c(x)$$, such that
 * $$y(x) = c(x) y_1(x)$$

4. Find the 2nd homogeneous solution $$y_2(x)$$

Repeat all of the steps for Euler L2-ODE-CC

Part 1.1
Assuming trial solution
 * $$y = x^r$$ where $$r$$ is a constant
 * $$\Rightarrow y' = r x^{r-1}$$
 * $$\Rightarrow y'' = r(r-1) x^{r-2}$$

Substituting into ($$)
 * $$a_0 r (r-1) x^r + a_1 r x^r + a_2 = 0$$
 * $$\Rightarrow \left[ a_0 r^2 + (a_1 - a_0) r + a_2 \right]x^r = 0$$

Since $$x^r \neq o$$
 * $$\Rightarrow a_0 r^2 + (a_1 - a_0) r + a_2 = 0$$

Expanding ($$)

Comparing coefficients of ($$) and ($$)
 * $$ \frac{a_2}{a_0} = 25 $$ and $$ \frac{a_1}{a_0} - 1 = -10$$

Taking $$a_0 = k$$, we have

Part 1.2
Since the characteristics equation is given below
 * $$(r - 5)^2 = 0$$

The roots of the characteristics equation are at $$r=5$$

So the homogeneous solution is given by
 * $$y_1(x) = x^r = x^5$$

Part 1.3
Assuming $$y(x) = c(x)y_1(x) = c(x)x^5$$
 * $$\Rightarrow y'(x) = c'(x)x^5 + 5 c(x)x^{5-1} = c'(x)x^5 + 5 c(x)x^4$$
 * $$\Rightarrow y(x) = c(x)x^5 + 10 c'(x)x^4 + 20 c(x)x^3$$

Substituting ($$) into ($$)
 * $$k x^2 y'' - 9k x y' + 25k y = 0$$
 * $$\Rightarrow x^2 y'' - 9 x y' + 25 y = 0$$

Using the expressions for $$y''(x), y'(x), y(x)$$


 * $$ c''(x)x^7 + 10 c'(x)x^6 + \cancel{20 c(x)x^5} - 9 c'(x)x^6 - \cancel{45 c(x)x^5} + \cancel{25 c(x)x^5} = 0$$
 * $$\Rightarrow c''(x)x^7 + c'(x)x^6 = 0$$

Substituting $$g(x) = c'(x)$$ and considering the fact that $$x^6 \neq 0$$
 * $$\Rightarrow g'(x)x + g(x) = 0$$
 * $$\Rightarrow \frac {g'(x)} {g(x)} = -\frac {1} {x}$$

Integrating w.r.t. $$x$$
 * $$\Rightarrow log(g(x)) = -log(x)+k_1$$
 * $$\Rightarrow log(g(x)) + log(x) = k_1$$
 * $$\Rightarrow g(x)x = e^{k_1} = k_2$$
 * $$\Rightarrow g(x) = \frac {k_2}{x}$$
 * $$\Rightarrow c'(x) = \frac {k_2}{x}$$

Integrating w.r.t. $$x$$
 * $$\Rightarrow c(x) = k_2 \log(x) + k_3$$


 * $$\Rightarrow y(x) = (k_2 \log(x) + k_3) x^5$$

Part 1.4
The solution, which is presented in ($$), is linear combination of two terms, $$y_1(x) = x^5$$ and $$\log(x) x^5$$

So the second homogeneous solution is $$y_2(x) = \log(x) x^5$$

Part 2.1
Let us assume the trial solution
 * $$y(x) = e^{rx}$$
 * $$\Rightarrow y'(x) = re^{rx}$$
 * $$\Rightarrow y''(x) = r^2 e^{rx}$$

Substituting into ($$)
 * $$b_0 r^2 e^{rx} + b_1 re^{rx} + b_2 e^{rx} = 0$$
 * $$\Rightarrow \left(b_0 r^2 + b_1 r+ b_2 \right) e^{rx} = 0$$

Since $$e^{rx} \neq 0$$
 * $$\Rightarrow b_0 r^2 + b_1 r+ b_2 = 0$$

Expanding ($$)

Comparing coefficients of ($$) and ($$)
 * $$ \frac{b_2}{b_0} = 25 $$ and $$ \frac{b_1}{b_0} = -10$$

Taking $$b_0 = k$$, we have

Part 2.2
Since the characteristics equation is given below
 * $$(r - 5)^2 = 0$$

The roots of the characteristics equation are at $$r=5$$

So the homogeneous solution is given by
 * $$y_1(x) = e^{rx} = e^{5x}$$

Part 2.3
Assuming $$y(x) = c(x)y_1(x) = c(x)e^{5x}$$
 * $$\Rightarrow y'(x) = c'(x)e^{5x} + 5 c(x)e^{5x} $$
 * $$\Rightarrow y(x) = c(x)e^{5x} + 10 c'(x)e^{5x} + 25 c(x)e^{5x}$$

Substituting ($$) into ($$)
 * $$k y'' - 10 k y' + 25k y = 0$$
 * $$\Rightarrow y'' - 10 y' + 25 y = 0$$

Using the expressions for $$y''(x), y'(x), y(x)$$


 * $$ c''(x)e^{5x} + \cancel{10 c'(x)e^{5x}} + \cancel{25 c(x)e^{5x}} - \cancel{10 c'(x)e^{5x}} - \cancel{50 c(x)e^{5x}} + \cancel{25 c(x)e^{5x}} = 0$$
 * $$\Rightarrow c''(x)e^{5x} = 0$$

Since $$e^{5x} \neq 0$$
 * $$\Rightarrow c''(x) = 0$$

Integrating twice w.r.t. $$x$$
 * $$\Rightarrow c(x) = k_1x + k_2$$


 * $$\Rightarrow y(x) = (k_1x + k_2) e^{5x}$$

Part 2.4
The solution, which is presented in ($$), is linear combination of two terms, $$y_1(x) = e^{5x}$$ and $$xe^{5x}$$

So the second homogeneous solution is $$y_2(x) = xe^{5x}$$

Pavel Bhowmik (talk) 21:31, 13 November 2012 (UTC)

=Problem 6.9* =

Given:
The following L2-ODE-VC

Find:
A valid homogeneous solution and call it $$ u_1(x) $$.

Find the 2nd homogeneous solution $$ u_2(x) $$ using variation of parameters method and compare it to $$ e^{xr_2(x)} $$.

Solution:
Let us assume the trial solution
 * $$y(x) = e^{rx}$$
 * $$\Rightarrow y'(x) = re^{rx}$$
 * $$\Rightarrow y''(x) = r^2 e^{rx}$$

Substituting into ($$)
 * $$(x-1) r^2 e^{rx} - x re^{rx} + e^{rx} = 0$$
 * $$\Rightarrow \left( (x-1) r^2 - x r + 1 \right) e^{rx} = 0$$

Since $$e^{rx} \neq 0$$
 * $$\Rightarrow (x-1) r^2 - x r + 1 = 0$$
 * $$\Rightarrow (x-1) r^2 - (x - 1) r - r + 1  = 0$$
 * $$\Rightarrow (x - 1) r (r - 1) - (r - 1 ) = 0$$
 * $$\Rightarrow ((x - 1) r - 1 )(r - 1 ) = 0$$

The roots of $$r$$ are
 * $$r=1, r=\frac{1}{x-1}$$

Taking the root $$r=1$$, the homogeneous solution is $$y_1(x) = e^x$$

Assuming $$y(x) = c(x)y_1(x) = c(x)e^x$$
 * $$\Rightarrow y'(x) = c'(x)e^x + c(x)e^x $$
 * $$\Rightarrow y(x) = c(x)e^x + 2c'(x)e^x + c(x)e^x $$

Substituting into ($$)
 * $$(x - 1) (c''(x)e^x + 2c'(x)e^x + c(x)e^x) - x (c'(x)e^x + c(x)e^x) + c(x)e^x = 0$$

Since $$e^x \neq 0$$
 * $$(x - 1) (c''(x) + 2c'(x) + c(x)) - x (c'(x) + c(x)) + c(x) = 0$$
 * $$\Rightarrow xc(x) + 2xc'(x) + \cancel{xc(x)} - c(x) - 2c'(x) - \cancel{c(x)} - x c'(x) - \cancel{x c(x)} + \cancel{c(x)} = 0 $$

Substituting $$ c'(x) = p(x) $$
 * $$\Rightarrow c''(x) = p'(x) $$


 * $$\Rightarrow xp'(x) + xp(x) - p'(x) - 2p(x) = 0 $$
 * $$\Rightarrow xp'(x) - p'(x) = 2p(x) - xp(x) $$
 * $$\Rightarrow \frac {p'(x)}{p(x)} = \frac {1} {x - 1} - 1 $$

Integrating w.r.t. $$x$$
 * $$\Rightarrow \log (p(x)) = \log (x - 1) - x + k_1$$
 * $$\Rightarrow p(x) = (x - 1) e^{- x + k_1} = (x - 1) e^{- x }e^{ k_1} = k_2(x - 1) e^{- x }$$
 * $$\Rightarrow c'(x) = k_2 x e^{- x } - k_2 e^{- x }$$
 * $$\Rightarrow c'(x) = k_2 x e^{- x } - k_2 e^{- x }$$

Integrating w.r.t. $$x$$
 * $$\Rightarrow c(x) = k_2 \int^x x e^{- x } \mathrm{d} x+ k_2 e^{- x } + k_3$$

Integrating by parts
 * $$\Rightarrow c(x) = k_2 x \int^x e^{- x } \mathrm{d} x - k_2 \int^x \left[ \frac{\mathrm{d}}{\mathrm{d}x}x \int^x e^{- x }\mathrm{d} x \right]\mathrm{d} x + k_2 e^{- x } + k_3$$


 * $$\Rightarrow c(x) = - k_2 x e^{- x } + k_2 \int^x e^{- x } \mathrm{d} x + k_2 e^{- x } + k_3$$


 * $$\Rightarrow c(x) = - k_2 x e^{- x } - \cancel{k_2 e^{- x }} + \cancel{k_2 e^{- x }} + k_3 = - k_2 x e^{- x } + k_3$$


 * $$\Rightarrow y(x) = c(x) e^x = (- k_2 x e^{- x } + k_3)e^x$$

The solution, which is presented in ($$), is linear combination of two terms, $$y_1(x) = e^{x}$$ and $$x$$

So the second homogeneous solution is $$y_2(x) = x$$, which is equal to $$e^{xr(x)}$$ only if
 * $$r(x) = \frac {\log{x}} {x}$$

Pavel Bhowmik (talk) 05:28, 14 November 2012 (UTC)