This equation is \(y'+2y=0\)

Its characteristic equation is \(\displaystyle{r}+{2}={0}\to{r}=-{2}\)

Therefore, the solution is \(\displaystyle{y}={C}{e}^{{-{{2}}}}{x}\)

Question

asked 2021-06-16

\(\displaystyle{2}⋅{\left(\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}\right)}+{2}{y}={0}\)

asked 2021-09-11

Find \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\) and \(\displaystyle{\frac{{{d}^{{2}}{y}}}{{{\left.{d}{x}\right.}^{{2}}}}}.{x}={e}^{{t}},{y}={t}{e}^{{-{t}}}\). For which values of t is the curve concave upward?

asked 2021-05-03

Find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}.x=e^t,y=te^{-t}\). For which values of t is the curve concave upward?

asked 2021-05-17

Compute \(\triangle y\) and dy for the given values of x and \(dx=\triangle x\)

\(y=x^2-4x, x=3 , \triangle x =0,5\)

\(\triangle y=???\)

dy=?

\(y=x^2-4x, x=3 , \triangle x =0,5\)

\(\triangle y=???\)

dy=?