Question

solve for initial problem of:
dy/dx: x(2+x^2)^4, y(0)=0

Answer 1

just integrate the function. the constant of integration can be taken as y(0)=0.

Answer 2

I did try to ingrate the function but I think I did it wrong or something

Answer 3

take u=2+x^2
then,
du=2xdx
so,\[\frac{1}{2}\int\limits_{}^{}u^4 du\]go from there

Answer 4

But, then y(0) would not equal zero...so perhaps I am misunderstanding the question?

Answer 5

I got it, thanks

Answer 6

Unless it should read\[y_0=0\]?

Answer 7

can you help me on something else?

Answer 8

no, it's y(0)=0

Answer 9

that makes no sense. y(0)=32/10 not 0

Answer 10

idk, the solution is y=1/10(2+x^2)^5-16/5

Answer 11

so I guess the answer is:\[y(x)=\frac{1}{10}(2+x^2)^5-\frac{32}{10}\]

Answer 12

lol...yeah same thing.

Answer 13

how did you get that?? that is right

Answer 14

got it :)

Answer 15

can you walk me thru that?

Answer 16

what did you get for the integral?

Answer 17

should I multiply out first before integral?

Answer 18

you get\[u=\frac{1}{10}u^5+u_0\]for the integral

Answer 19

then sub back in u=2+x^2
\[y(x)=\frac{1}{10}(2+x^2)^5+y_0\]Since we know y(0)=0...we can solve for y_0

Answer 20

oh ic, thanks

Answer 21

np

Answer 22

can you help me on something else?

Answer 23

ok. post it as a new question tho

Answer 24

thanks bye in here