# unique solution.

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• May 26th 2012, 03:00 AM
saravananbs
unique solution.
dy/dx = y1/3 and y(0)=0

as unique solution. is it true?

what about dy/dx= |y|1/3 and y(0)=0

thank u
• May 26th 2012, 07:38 AM
Prove It
Re: unique solution.
\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} &= y^{\frac{1}{3}} \\ y^{-\frac{1}{3}}\,\frac{dy}{dx} &= 1 \\ \int{y^{-\frac{1}{3}}\,\frac{dy}{dx}\,dx} &= \int{1\,dx} \\ \int{y^{-\frac{1}{3}}\,dy} &= x + C_1 \\ \frac{3}{2}y^{\frac{2}{3}} + C_2 &= x + C_1 \\ \frac{3}{2}y^{\frac{2}{3}} &= x + C_1 - C_2 \\ y^{\frac{2}{3}} &= \frac{2}{3}x + \frac{2}{3}C_1 - \frac{2}{3}C_2 \\ y^{\frac{2}{3}} &= \frac{2}{3}x + C \textrm{ where }C = \frac{2}{3}C_1 - \frac{2}{3}C_2 \\ |y| &= \left(\frac{2}{3}x + C\right)^{\frac{3}{2}} \\ |0| &= \left(\frac{2}{3}\cdot 0 + C\right)^{\frac{3}{2}} \textrm{ after substituting the point } y(0) = 0 \\ 0 &= C^{\frac{3}{2}} \\ C &= 0 \\ |y| &= \left(\frac{2}{3}x\right)^{\frac{3}{2}} \\ y &= \pm \left(\frac{2}{3}x\right)^{\frac{3}{2}} \end{align*}

So no, the solution is not unique.