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]) ?
\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]) ?