User:Egm6321.f12.team4.mishra/Homework2 R2.4 & R2.5

Statement

 * To show that $$ y_{H}^{1}(x) $$ and $$ y_{H}^{2}(x) $$ are linearly independent
 * where,
 * $$ y_{H}^{1}(x) = x $$:
 * $$ y_{H}^{2}(x) = \frac{x}{2}log(\frac{1+x}{1-x})-1 $$
 * $$ y_{H}^{2}(x) = \frac{x}{2}log(\frac{1+x}{1-x})-1 $$

Solution
From the definition of linear independence, we have to show that for all x,


 * $$ y_{H}^{1}(x) \neq \alpha y_{H}^2(x)$$ for some $$ \alpha (\neq 0) $$


 * Since the above should hold for all value of x, let us assume a value of x equal to be 0,

Evaluating $$ y_{H}^{1}(x) $$ & $$ y_{H}^{2}(x) $$ at x=0. $$ y_{H}^{1}(x) = 0 $$  Eq. (2.4.1) $$ y_{H}^{2}(x) = \frac {0}{2}log(\frac{1+0}{1-0})-1 = -1 $$  Eq. (2.4.2)


 * For linear dependence,
 * $$ y_{H}^{1}(x) $$ = $$\alpha y_{H}^{2}(x) $$,
 * $$0 = -\alpha $$

We can easily see that for any value of $$\alpha (\neq 0)$$, the above will not hold true. This proves the linear independence of the two functions.

Author and References
Solved and typed by -- Pushkar Mishra

Statement
Consider the following function $$ \phi (x,y)=x^{2}y^{3/2}+log(x^3y^2)=k $$

Find  $$ G(y',y,x) = \frac{d}{dx} \phi(x,y)=0 $$

Solution
We start by differentiating $$ \phi(x,y) $$ with respect to x. Since the function is a product of two functions dependent on x, we use differentiation by parts.

$$ (\frac{d}{dx}x^2)y^{3/2}+x^2(\frac{d}{dx}y^{3/2})+\frac{1}{x^3y^2}((\frac{d}{dx}x^3)y^2+x^3\frac{d}{dx}y^2) $$  Eq. (2.5.1)

It follows that,

$$ 2xy^{2/3} + x^2\frac{3}{2}y^{1/2}\frac{dy}{dx} + \frac{1}{x^{3}y^{2}} (3x^{2}y^{2}+2x^{3}y\frac{dy}{dx}) =0 $$  Eq. (2.5.2)

Simplifying the equation,

$$ 2xy^{2/3} + \frac{3}{2}y^{1/2}x^2\frac{dy}{dx} + \frac{3}{x}+ \frac{2}{x}\frac{dy}{dx} =0 $$  Eq. (2.5.3)

It can be seen that the above is a function of x,y,y'.Arranging the terms we get,

$$ G(x,y,y')=(2xy^{2/3} + \frac{3}{x})+ (\frac{3}{2}y^{1/2}x^2 + \frac{2}{x})y' =0 $$ Eq. (2.5.4)

which can be written as ,

$$ G(x,y,y') = M(x,y) + N(x,y)y'=0 $$


 * where ,


 * $$ M(x,y) = 2xy^{2/3} + \frac{3}{x} $$ and


 * $$ N(x,y) = \frac{3}{2}y^{1/2}x^2 + \frac{2}{x} $$

Since the highest derivative of y in the equal is 1, G(y',y,x) is a 1st order ODE in y.

Author and References
Solved and typed by -- Pushkar Mishra