Solution of $\frac{dy}{dx}=-y^\frac{1}{3} $ and $y(0)=0 $.

Asked Oct 02 '22 at 18:30

Active Oct 02 '22 at 18:30

Viewed 115 times

$\frac{dy}{dx}=-y^\frac{1}{3} $ and $y(0)=0 $.

Then this Initial value problem has

a) unique solution

b) infinity many solution

c) No solution

My apporach: Take $y= (-\frac{2}{3}(x-a))^\frac{3}{2}$ if $x\leq a$

And $y=0$ if $x\geq a$ Where a is any negative real no. It satisfy Initial condition. It have infinitely many solution for Infinitely many choice of a. So IVP has infinitely many solution.

Is my approach correct?

asked Oct 02 '22 at 18:30

Santosh Negi

1

A similar question was asked recently: https://math.stackexchange.com/q/4543015/42969 – Martin R Oct 02 '22 at 18:40
Thankyou , i was not able to find that question – Santosh Negi Oct 02 '22 at 18:45
$dx/dy=-y^{-1/3}$ so $x=-\frac32 y^{2/3}+C.$ So $y=(C_1-\frac23 x)^{3/2}.$ – Thomas Andrews Oct 02 '22 at 19:09

Solution of $\frac{dy}{dx}=-y^\frac{1}{3} $ and $y(0)=0 $.

0 Answers0