From integral equation to normal equation

[ManishR]

Consider

$$\frac{d^{2}y}{dx^{2}}=k$$--------------[0]

and

$$\frac{dy}{dx}=y_{a}$$

Then

$$y_{f}-y_{i}=\frac{k}{2}(x_{f}^{2}-x_{i}^{2})$$-----------[1]

$$y_{af}-y_{ai}=k(x_{f}-x_{i})$$-------------[2]

is equation 1 and 2 correct ? if no, then what is the correct solution

if yes

$$\Rightarrow\frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}=(positive/negative)$$

or more correctly

$$\Rightarrow\frac{\frac{d^{2}y}{dx^{2}}}{\left|\frac{d^{2}y}{dx^{2}}\right|}=\frac{\frac{dy}{dx}}{\left|\frac{dy}{dx}\right|}=\frac{k}{\left|k\right|}$$

assumming

$$x_{f}>x_{i}$$

but that's not always true for example its not true for this

$$y=2x-x^{2}$$-------------------[4]

where

$$-1\leq x\leq1$$

$$\Rightarrow\frac{dy}{dx}=2-2x=positive$$

$$\Rightarrow\frac{d^{2}y}{dx^{2}}=-2=negative$$

so how to derive from integral equation (like [0]) to normal equation(like [4]) ?

 

[jjmontero9]

This can be done using the http://en.wikipedia.org/wiki/Laplace_transform" , which is really useful for solving differential equations. First you apply the direct transformation to [0], then solve for y and then apply the inverse laplace transform to find an equation like [4].

An example on http://en.wikipedia.org/wiki/Laplace_transform#Example_.231:_Solving_a_differential_equation"

 

[ManishR]

This can be done using the http://en.wikipedia.org/wiki/Laplace_transform" , which is really useful for solving differential equations. First you apply the direct transformation to [0], then solve for y and then apply the inverse laplace transform to find an equation like [4].

An example on http://en.wikipedia.org/wiki/Laplace_transform#Example_.231:_Solving_a_differential_equation"

Its not a differential equation.
$$\frac{dk}{dx}=\frac{dk}{dy}=0$$

 

[dextercioby]

Your [1] doesn't really follow from [0]. There's also a linear term in "x" in the RHS.

 

[ManishR]

thanks bigubau
$$y_{f}-y_{i}=y_{ai}(x_{f}-x_{i})+\frac{k}{2}(x_{f}^{2}-x_{i}^{2})$$ -------[1]

 

[ManishR]

Your [1] doesn't really follow from [0]. There's also a linear term in "x" in the RHS.

could you explain how you got x ?
i still cannot see how equation [1] is wrong ?