Problem: Solve the initial value problem
$$y′ = \frac{1+3x^2}{3y^2 −6y},\quad y(0) = 1$$
and determine the interval in which the solution is valid. Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.
Hi! I have this problem from one of my homework's and I am stuck on the part of determining the interval. I have already separated the variables and found C using the initial condition and got:
y^3 - 3y^2 = x + x^3 -2.
Now I need to determine the interval for the solution. I tried to understand the hint but can't really figure what they mean by looking for points where the integral curve has a vertical tangent. Do I accomplish this by setting:
dx/dy = 0
Thanks for the help!
Best Answer
From the original DE, $y'$ is undefined (vertical tangent) when $y\in\{0,2\}$ as the denominator vanishes then. These values can be solved for the corresponding x-values through the equation for the solution
$$y^3-3y^2=x^3+x-2 \tag{1}$$
For $y=0$ $$x^3+x-2=0 \implies (x-1)(x^2+x+2)=0 \implies x=1,x=\frac{-1\pm i\sqrt7}{2}$$ with the only real solution $x=1$.
For $y=2$ $$x^3+x+2=0 \implies (x+1)(x^2-x+2)=0 \implies x=-1,x=\frac{1\pm i\sqrt7}{2}$$ with the only real solution $x=-1$.
Since the initial condition is between these two values, one could argue that it, and hence the solution as given in (1), apply only over the open interval $(-1,1)$.
Remark
If it is known that for $y=f(x)$ the function $f$ is continuous at $-1$ and at $1$, then these can be used as initial values for the solutions on $(-\infty,-1]$ and $[1,\infty)$ but this information is not given.