From the existness and uniqueness Theorem,the initial value problem
$$y'=3x(y-1)^{1/3} , y(3)=-7$$
has a unique solution on some open interval that contains $x=3$. Find the solution and determine
the largest open interval on which it’s unique.
What i tried,
First i tried to solve the equation by the seperable equation method to get,$$y=1+(x^2-5)^{1.5}$$. Then from here i used the existence and uniquness theorem to calculate $f(x,y)$ and $f_{y}$. From the calculations,when $y$ not equals to $1$, $f_{y}$ will be continuous, hence there will be a unique solution when $y$ not equals to $1$, according to the wxistness and uniquness theorem. However im stuck from here onwards as to finding the interval of validaty. Is my working correct. Could anyone explain. Thanks
Best Answer
once you have $y = 1 - (x^2 -5)^{3/2}$, you can see that $\sqrt 5 \le x < \infty$ is the maximal interval of existence.