User:Egm6321.f10.team4/HW 5

= Problem 1 - Fifth Derivative of $$\displaystyle y(x(t)) $$=

Given
$$\displaystyle x = e^t $$

$$\displaystyle y _t =  \frac {dy}{dt}  =  \frac {\partial y}{\partial x}  \frac {\partial x}{\partial t} $$

Find
$$\displaystyle y_{xxxxx} $$

Solution
The process is fairly straightforward, as we must first come up with an expression for $$\displaystyle y_x $$, and then take successive derivatives until $$\displaystyle y_{xxxxx} $$ is reached.

$$\displaystyle \frac {dy}{dt} =  y_t $$

$$\displaystyle \frac {\partial y}{\partial x} =  y_x $$

$$\displaystyle \frac {\partial x}{\partial t} =  e^t $$

Therefore the equation

$$\displaystyle y_t =  y_x e^t $$

can be solved for $$\displaystyle y_x $$

$$\displaystyle y_x =  e^{-t} y_t $$

From this point, we simply take each derivative until $$\displaystyle y_{xxxxx} $$ is found.

$$\displaystyle y_{xx} =  \frac {\partial}{\partial x}  \frac {dy}{dx}  =  e^{-t} (-e^{-t} y_t + e^{-t} y_{tt}) $$

$$\displaystyle y_{xx} =  e^{-2t} (y_{tt} - y_t) $$

$$\displaystyle y_{xxx} =  \frac {\partial}{\partial x} (y_{xx}) $$

$$\displaystyle y_{xxx} =  e^{-t} (e^{-2t} y_{ttt} - 2e^{-2t} y_{tt} - e^{-2t} y_{tt} + 2e^{-2t} y_t) $$

$$\displaystyle y_{xxx} =  e^{-3t} (y_{ttt} - 3y_{tt} + 2y_t) $$

$$\displaystyle y_{xxxx} =  \frac {\partial}{\partial x} (y_{xxx}) $$

$$\displaystyle y_{xxxx} =  e^{-t} (e^{-3t} y_{tttt} - 3e^{-3t} y_{ttt} - 3e^{-3t} y_{ttt} + 9e^{-3t} y_{tt} + 2e^{-3t} y_{tt} - 6e^{-3t} y_t) $$

$$\displaystyle y_{xxxx} =  e^{-4t} (y_{tttt} - 6 y_{ttt} + 11 y_{tt} - 6 y_t) $$

$$\displaystyle y_{xxxxx} =  \frac {\partial}{\partial x} (y_{xxxx}) $$

$$\displaystyle y_{xxxxx} =  e^{-t} (e^{-4t} y_{ttttt} - 4e^{-4t} t_{tttt} - 6e^{-4t} y_{tttt} + 24e^{-4t} y_{ttt} + 11e^{-4t} y_{ttt} - 44e^{-4t} y_{tt} - 6e^{-4t} y_{tt} + 24e^{-4t} y_t) $$

$$\displaystyle y_{xxxxx} =  e^{-5t} (y_{ttttt} - 10 y_{tttt} + 35 y_{ttt} - 50 y_{tt} + 24 y_t) $$

Egm6321.f10.team4.osentowski 18:08, 20 October 2010 (UTC) Egm6321.f10.team4.osentowski

= Problem 2 - Method of Trial Solution =

Given
$$\displaystyle x^2 y'' - 2x y' + 2y =  0 $$

$$\displaystyle y(1) =  -2 $$

$$\displaystyle y(2) =  5 $$

Find
$$\displaystyle y(x) $$, satisfying the boundary conditions.

Solution
Using the Method of Trial Solutions, our potential solution should be of the form

$$\displaystyle y =  x^r $$

Therefore

$$\displaystyle y' =  r x^{r - 1} $$

$$\displaystyle y'' =  r (r - 1) x^{r - 2} $$

Substituting these values into our given equation (2-1)

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

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

This simplifies to the characteristic equation

$$\displaystyle r^2 - 3r + 2 = 0 $$

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

Our possible values for $$\displaystyle r $$ are

$$\displaystyle r_2 = 2 $$

$$\displaystyle r_1 = 1 $$

Since this is a linear equation, the solution will be a linear combination of the possible values, taking the form

$$\displaystyle y(x) =  c_1 x^{r_1} + c_2 x^{r_2} $$

$$\displaystyle y(x) =  c_1 x + c_2 x^2 $$

Using our boundary conditions, a simple system of equations can be developed to determine the values of $$\displaystyle c_1 $$ and   $$\displaystyle c_2 $$

$$\displaystyle 5 =  c_1 (2) + c_2 (2^2) $$

$$\displaystyle -2 = c_1 (1) + c_2 (1^2) $$

Evaluating the system gives the values

$$\displaystyle c_1 = \frac {-13}{2} $$

$$\displaystyle c_2 = \frac {9}{2} $$

Our final solution for $$\displaystyle y(x) $$ is therefore

$$\displaystyle y(x) =  \frac {-13}{2} x + \frac {9}{2} x^2 $$

Egm6321.f10.team4.osentowski 18:08, 20 October 2010 (UTC) Egm6321.f10.team4.osentowski