Both y = 2e^t and y = e^t are solutions of the differential equation dy/dt = y. Since
this equation is autonomous the graph of y=2e^t can be obtained from the graph of
y=e^t by horizontal translation. By how much does one have to translate the graph
of y=e^t horizontally to obtain the graph of y=2e^t?
Does this question make sense?

Question

Both y = 2e^t and y = e^t are solutions of the differential equation dy/dt = y. Since
this equation is autonomous the graph of y=2e^t can be obtained from the graph of
y=e^t by horizontal translation. By how much does one have to translate the graph
of y=e^t horizontally to obtain the graph of y=2e^t?
Does this question make sense?

turingtest · Answer

I'm thinking ln2$$\large e^{t+\ln2}=2e^t$$...makes sense to me at least :D

anonymous · Answer

so what would the horizontal translation be?

turingtest · Answer

ln2 units to the left

turingtest · Answer

we are changing$$f(t)\to f(t+\ln2)$$which is a translation of ln2 units left

anonymous · Answer

do you know what the answer to this is?
Find the general solution to
(b) dy/dt = t^2y^3

i think i have the right answer but want to be sure.

turingtest · Answer

what did you get?

anonymous · Answer

i got $$y=\pm \sqrt{3/2t^3 + c}$$

turingtest · Answer

I think you lost a minus sign, no?

anonymous · Answer

theres a very good chance .. what should the asnwer look like?

turingtest · Answer

$$\int\frac{dy}{y^3}=\int t^2dt$$$$-\frac1{2y^2}=\frac{t^3}3+C\to-\frac3{2t^3+C}=y^2$$$$y=\pm\sqrt{-\frac3{2t^3+C}}$$which would change the domain

anonymous · Answer

thank you!