ex:$$y=mx+c$$ $$\frac{dy}{dx}=m$$ $$y=\frac{dy}{dx}.x +c$$ The line equation has 2 arbitrary constants..it implies the DE should be second order...but it isnt and c isnt eliminated too...How do i eliminate c?
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$\begingroup$ the DE is indeed second order, its $y'' = 0$. The equation $\frac{dy}{dx} = m$ is not the DE satisfied by every function of the form $mx+c$. This is only true if $m$ is a fixed number, and then yes, its first order $\endgroup$– Calvin KhorNov 7, 2019 at 8:57
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$$y=mx+c\quad\implies\quad \begin{cases}\frac{dy}{dx}=m\\ \frac{d^2y}{dx^2}=0\end{cases}$$ Thus the second order ODE is very simple : $$ \frac{d^2y}{dx^2}=0$$ In solving it you check that the general solution is $\quad y=mx+c\quad$ as expected.
answered Nov 7, 2019 at 9:20