Question

Solve this equation

Answer 1

\[\Large\begin{array}{rclr}
\frac{dy}{dx}&=&e^{-x}y\\
\frac1ydy&=&e^{-x}dx&\mbox{(STEP 1)}\\
\int\frac1ydy&=&\int e^{-x}dx&\mbox{(STEP 2)}\\
\ln y&=&-e^{-x}+C&\mbox{(STEP 3})\\
\end{array}\]
Do you understand all? If not, which step do you not understand? (I need this information to help you)

Answer 2

yes, i got the same, isit i just need to sub in the initial condition to find C and then solve for y(2)? just need to check(:

Answer 3

Yes :)

Answer 4

isit c= 1 -ln8.7
hm ln y= -e^-2 +(1-ln8.2) ???

Answer 5

I think c=1+ln8.7

Answer 6

isnt it y(0)=-8.7?

Answer 7

wait....

Answer 8

so \(\ln-8.7=-1+C\)?

Answer 9

But one cannot apply logarithm to a negative number...

Answer 10

what is the question is wrong

Answer 11

yeah.. so it must be positive?

Answer 12

what if the question is wrong

Answer 13

it can be.. btw if it was positive. how do you get y value from the above eqn?

Answer 14

\[\Large e^{\ln y}=y\]

Answer 15

okay thanks!(:

Answer 16

can you help me with another one?

Answer 17

sorry I don't know how to do :/