## anonymous 4 years ago 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?

1. TuringTest

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

2. anonymous

so what would the horizontal translation be?

3. TuringTest

ln2 units to the left

4. TuringTest

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

5. anonymous

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.

6. TuringTest

what did you get?

7. anonymous

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

8. TuringTest

I think you lost a minus sign, no?

9. anonymous

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

10. TuringTest

$\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

11. anonymous

thank you!