User:Egm6321.f10.team5.abc/HW2

Problem 3
Plot $$y_H^1(x)$$ & $$y_H^2(x)$$ and show that $$y_H^1(x) \ne \alpha y_H^2(x)$$ for any $$x$$ and any $$\alpha$$ that belong to $$\mathbb{R}$$ where $$y_H^1(x) = x$$ and $$y_H^2(x) = \frac x 2 \log \left(\frac {1+x} {1-x}\right) - 1$$

Solution 3


In the figure inserted above, the points of $$y_H^1(x)$$ are marked as "x" and the points of $$y_H^2(x)$$ are maked as "o".

To show that $$y_H^1(x)$$ and $$y_H^2(x)$$ are linearly independent, we use Wronskian. $$W(y_H^1,y_H^2)(x) = \left| \begin{matrix} {y_H^1} & {y_H^2} \\ {y_H^1}' & {y_H^2}'\end{matrix} \right| = \left| \begin{matrix} x & {\frac x 2 \log \frac {1+x} {1-x} - 1} \\ 1 & {\frac x {1-x^2} + \frac 1 2 \log \frac {x+1} {x-1}} \end{matrix} \right| \qquad\qquad\qquad\qquad\qquad(3.1)$$ $$= x (\frac x {1-x^2} + \frac 1 2 \log \frac {x+1} {x-1}) - (\frac x 2 \log \frac {1+x} {1-x} - 1) 1 \ne 0 \qquad\qquad\qquad\qquad\qquad(3.2)$$ Therefore we show that $$y_H^1(x)$$ and $$y_H^2(x)$$ are linearly independent.

Problem 4
Find $$F(x,y,y') = \frac {d\phi} {dx} (x,y)$$ where $$\phi(x,y) = x^2 y^{\frac 3 2} + \log (x^3 y^2) = k$$ and verify that $$F$$ is exact N1_ODE and invent 3 more.

Solution 4
First, we derive $$\frac {d\phi} {dx} (x,y)$$. $$F(x,y,y') = x y^{\frac 3 2} + \frac 1 {x^3 y^2} 3x^2 y^2 = x y^{\frac 3 2} + \frac 3 x \qquad\qquad\qquad\qquad\qquad(4.1)$$ The definition of exact N1_ODE for $$F$$ requires a function $$\phi(x,y)$$ such that $$F = \frac {d\phi} {dx}$$ to exist. Since we have the function $$\phi(x,y)$$, it is verified that F is exact N1_ODE. The following three functions are also exact N1_ODE.

1. $$F(x,y,y') = 2x$$. Then, we have a function $$\phi(x,y) = x^2$$ such that $$F = \frac {d\phi} {dx}$$

2. $$F(x,y,y') = \frac 1 x$$. Then, we have a function $$\phi(x,y) = \log(x)$$ such that $$F = \frac {d\phi} {dx}$$

3. $$F(x,y,y') = 1$$. Then, we have a function $$\phi(x,y) = x$$ such that $$F = \frac {d\phi} {dx}$$

Problem 9
Use $$\bar b(x) c(y) y' + a(x) \bar c(y) = 0$$ to find a N1_ODE that is either exact or can be made exact where $$a(x) = \sin x^3$$, $$b(x) = \cos x$$, and $$c(y) = \exp(2y)$$. Then find $$\phi(x,y) = k$$.

Solution 9
First, calculate $$\bar b(x)$$ and $$\bar c(x)$$. $$\bar b(x) = \int^x b(s) ds = \int^x \cos s \ ds = \sin x \qquad\qquad\qquad\qquad\qquad(9.1)$$ $$\bar c(x) = \int^y c(y) ds = \int^y \exp(2s) ds = \frac 1 2 \exp(2y) \qquad\qquad\qquad\qquad\qquad(9.2)$$ We express the equation completely. $$\sin x \ \exp(2y) y' + \sin x^3 \ \frac 1 2 \exp(2y) = 0 \qquad\qquad\qquad\qquad\qquad(9.3)$$ Dividing the both sides by $$\exp(2y)$$, we get $$\sin x \ y' + \sin x^3 \ \frac 1 2 = 0 \qquad\qquad\qquad\qquad\qquad(9.4)$$ To determine if the equation is exact or not $$\frac {\partial M} {\partial y} = 0 \qquad\qquad\qquad\qquad\qquad(9.5)$$ $$\frac {\partial N} {\partial x} = \cos x \ y' \qquad\qquad\qquad\qquad\qquad(9.6)$$ The eqauation is not exact since $$\frac {\partial M} {\partial y} \ne \frac {\partial N} {\partial x}$$, so it can be made exact. $$\sin x \ y' = - \sin x^3 \ \frac 1 2 \qquad\qquad\qquad\qquad\qquad(9.7)$$ Since the equation is the first-order derivative with respect to $$y$$, we need to take an integral with respect to y. $$y' = - \frac 1 2 \frac {\sin x^3} {\sin x} \qquad\qquad\qquad\qquad\qquad(9.8)$$ To express in the form of $$\phi(x,y) = k $$, we take the integration on each side. $$y = - \frac 1 2 \int \frac {\sin x^3} {\sin x} dx + k \qquad\qquad\qquad\qquad\qquad(9.9)$$ $$y + \frac 1 2 \int \frac {\sin x^3} {\sin x} dx = k \qquad\qquad\qquad\qquad\qquad(9.10)$$