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The differential equation $dy/dx = 60 (y^2) ^ {1/5} $; $x>0$ $y(0)=0$ has
(1) A unique solution.
(2) Two solutions.
(3) No solution.
(4) Infinite number of solutions.

After solving I get that solution will be $y^{7/5}=84x$ that is a unique solution. But the given answer is two solutions. Can anyone tell me please where my mistake is?

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1 Answer 1

up vote 1 down vote accepted

Presumably you observed that this is a separable equation, rewrote it as $$ y^{-\frac{2}{5}}dy = 60dx, $$ and solved, which is fine, unless $y$ is equal to $0$ somewhere (the point given in the problem $y(0) = 0$ is awkward since $x > 0$; they should have said $\lim_{x \to 0^+} y = 0$).

Investigating, we see that the solution where $y$ is constantly zero is another solution, hence there are two.

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in other words, your argument proved that there is a unique solution where $y$ is nonzero, but it said nothing about solutions where $y$ vanishes, since you implicitly assumed $y$ didn't vanish in performing separation of variables. –  user29743 Jan 25 '13 at 14:26
@poton: Check the Picard's theorem conditions to see why the solutions are not unique. +1 –  B. S. Jan 25 '13 at 14:26
In fact there are infinitely many solutions. For each $a\ge0$ the function $y_a$ defined as $y_a(x)=0$ if $0\le x\le a$, $y_a(x)=(36(x-a))^{5/3}$ if $x>a$ is a solution. –  Julián Aguirre Jan 25 '13 at 14:29
Note that the solution $y=0$ is a singular solution. –  B. S. Jan 25 '13 at 14:30
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