Question

I'm not really sure what to do on this problem.

Find the solution of the differential equation

y'=5x^4 e^(-x^5), such that y(0)=1

Answer 1

try to set \(u=-x^5\) first

Answer 2

yeah, i believe so.

Answer 3

its in the form
e^-u du

when
u = x^5
du = 5x^4

Answer 4

set \(u=x^5\)
so, we get \[y=\int e^{-u}du=-e^{-u}+C=-e^{-x^5}+C\]
set \(x=0\) then
\[y(0)=-e^{-0^5}+C\\
1 = -e^1+C\\
C=1+e\]

i think the result is obvius by now

Answer 5

hmm
\[e^{0^{5}} = e^{0} = 1\]

Answer 6

oh sorry :)

Answer 7

it must be C=2