Question

Dy/dx=cubertof y

Answer 1

cross multiply
\[\large \implies \frac{dy}{\sqrt[3]{y}} = dx\]

Answer 2

then change \[\frac{dy}{\sqrt[3] y}\]
into \[\large y^{-\frac 13 } dy\]
so the equation becomes
\[\Large \implies y^{-\frac 13} dy = dx\]
does that help?

Answer 3

...then integrate both sides :)

Answer 4

yes. exactly

Answer 5

-3/2y^-2/3=x

Answer 6

+c

Answer 7

Y=((-2x+1)/3)^-2/3

Answer 8

why negative again?

Answer 9

\[\LARGE \int y^{-\frac 13} dy \implies \frac{ y^{-\frac 13 + \frac 33}}{-\frac 13 + \frac 33}\]

Answer 10

.??

Answer 11

you integrated wrong

Answer 12

Yprime of y= -3/2x^-2/3 then I'd get x^-1/3

Answer 13

you integrated y^-(1/3) dy wrong

Answer 14

it's not -3/2 y^-2/3

Answer 15

??

Answer 16

remember power ryle

Answer 17

rule*

Answer 18

No

Answer 19

no?

Answer 20

no?

Answer 21

well you did it right...except for the negative sign

\[\huge \int y^{-\frac 13} dy \implies \frac{y^{-\frac 13 + 1}}{-\frac 13 + 1} \implies \frac{y^{\frac 23}}{\frac 23} \implies \frac{3y^{\frac 23}}2\]
got it now?

Answer 22

Yes

Answer 23

Y(0)=1
So what is final answer?

Answer 24

there's a condition that y(0) = 1?

Answer 25

Yes

Answer 26

well your equation is \[\huge \frac {3y^\frac 23}2 = x + c\]
if y(0) = 1
\[\huge \implies \frac{3(1)^{\frac 23}}2 = 0 +c\]
\[\huge \implies \frac 32 = c\]

Answer 27

so now substitute c = 3/2 into the original equation...
\[\huge \implies \frac{3 y^{\frac 23}}2 = x + \frac 32\]

Answer 28

Ty