From integral equation to normal equation

ManishR
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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]) ?
 
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jjmontero9 said:
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
 
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Your [1] doesn't really follow from [0]. There's also a linear term in "x" in the RHS.
 
thanks bigubau
y_{f}-y_{i}=y_{ai}(x_{f}-x_{i})+\frac{k}{2}(x_{f}^{2}-x_{i}^{2}) -------[1]
 
bigubau said:
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 ?
 
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