Question

The initial value problem \[\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 3}}{{\bf{x}}^{\frac{{\bf{2}}}{{\bf{3}}}}}{\bf{,}}\;{\bf{x(0) = 1}}\] has a unique solution in some open interval around t = 0.

Short answer

The given statement is true.

Step-by-step solution

Finding partial derivatives

Since \[\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 3}}{{\bf{x}}^{\frac{{\bf{2}}}{{\bf{3}}}}}\]

Then\[\frac{{\partial {\bf{f}}}}{{\partial {\bf{x}}}}{\bf{ = }}\frac{{\bf{2}}}{{\sqrt[{\bf{3}}]{{\bf{x}}}}}\]

Checking the final result

Apply the initial conditions\[x\left( 0 \right) = 1\]

\[\frac{{\partial f}}{{\partial x}}{\bf{ = }}\frac{{\bf{2}}}{{\bf{0}}}\]

The result is infinite.

The given function is discontinuous a x = 0. So the function is not continuous in a rectangle containing the point (0,1).

Therefore, the statement is True.