Question

determine by inspection at least two
solutions of the given first-order IV
y'= 3y^(2/3) y(0) = 0

Answer 1

looks like a trick question

Answer 2

if initial values are given, then there can exist oly one solution right ?

Answer 3

there are two... so the book says
they ended up making it dy/3y^(2/3) = dx

Answer 4

thats okay, but i dont see yet how two solutions can be there uhmm

Answer 5

lets dive in and see wat we get maybe..

Answer 6

next, integrate both sides

Answer 7

\(\large y'= 3y^{\frac{2}{3}} \)

Answer 8

its same as :

\(\large \frac{dy}{dx}= 3y^{\frac{2}{3}} \)

Answer 9

cross multiply

Answer 10

\(\large \frac{dy}{3y^{\frac{2}{3}} }= dx \)

Answer 11

integrate both side

Answer 12

\(\large \int \frac{dy}{3y^{\frac{2}{3}} }= \int dx \)

Answer 13

\(\large \frac{1}{3}\int y^{-\frac{2}{3}} dy= \int dx \)

Answer 14

\(\large \frac{1}{3} \frac{y^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}= x + c\)

Answer 15

simplify a bit

Answer 16

\(\large y^{\frac{1}{3}}= x + c\)

Answer 17

thats it?

Answer 18

if u plugin (0, 0), u wud get c = 0
so the solution wud be :

\(\large y^{\frac{1}{3}}= x \)

\(\large y= x^3 \)

Answer 19

thats the oly one final solution

Answer 20

you're talking about TWO solutions ?

Answer 21

*ur textbook

Answer 22

yeah it says at least two... so idk

Answer 23

uhmm u dont have answers ?

Answer 24

i have y = 0 and y = x^3