How many solutions does the differential equation have?

Apr 2017
87
1
India
The Initial Value Problem y′=y^2/3, y(0)=0 has ?

(a) No solution.

(b) Unique solution.

(c) Two solutions.

(d) Infinitely many solutions.

As I understood, one solution is y=0 itself and other solution comes on solving the differential equation as 3y^(1/3)= x+c, on putting the initial conditions, I get the solution as y=(x^3)/27.
Hence my conclusion is the answer is option c) that is only two solutions. Am I correct?
 
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SDK

Sep 2016
804
544
USA
You can use the two solutions you have found to construct infinitely many solutions so the answer is (d). Try it to construct these yourself.
 

skipjack

Forum Staff
Dec 2006
21,479
2,470
There are infinitely many piecewise-defined solutions.

One such solution is y = x³/27 for x < 0, y = 0 for x in [0, 1], and y = (x - 1)³/27 for x > 1.
 
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SDK

Sep 2016
804
544
USA
Better just solve the equation rather than applying theorems of existence of solution of a DE.
https://faculty.math.illinois.edu/~tyson/existence.pdf
The classical existence/uniqueness theorem does not apply to the OP example because the derivative in the example isn't a Lipschitz function. That is precisely why the solutions aren't unique. But the theorem you linked here doesn't say anything at all about uniqueness because it doesn't apply at all.
 
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