Where to Go Next for Uncertain Times

[franky2727]

question attached

 

[CompuChip]

You write
$$\int \frac{1}{(v + 1)^2} dv = \int \frac{1}{x} dx$$
and then on the next line
$$-v - 1 = \ln |v| + c$$

I think you made a writing error there, which leads to an insolvable equation.

 

[franky2727]

ah ok but if its lnx instead of v then I'm still on the right tracks?

 

[franky2727]

ok so where do i go from e^-v-1=KX ?

 

[CompuChip]

Yes, until that step you were fine.
Remember, you want the expression for v(x), and finally y(x) = x v(x).
So you have to solve v(x) from $e^{-v(x)-1} = k x$. Try taking logarithms on both sides.