Solve the given initial value problem and determine how the interval
in which the solution exists depends on the initial value $y_0$:
$$\frac {dy}{dx} + y^3 = 0,\;y(0)=y_0 $$
So far I've solved the DE and got that $y=\frac {1}{\sqrt {2t+2C}} $… I tried plugging in the initial value and got that$ C = \frac{1}{2y_0^2}$.
The answer to this problem is: $y=\frac{y0}{\sqrt{2ty_0^2+1}}$ if $y_0$ does not equal $0$. If $y_0$ does equal $0$, $\;y=0$.
I think I'm headed in the right direction but not sure because my constant value seems wrong – How did $y_0$ go to the numerator? Also, given the solution, how do you figure out how the interval depends on the initial value $y_0$?
Best Answer
$$ \frac {dy}{dx}=-y^3$$
$$ y^{-3} dy = -dx $$ $$\frac {-1}{2} y^{-2} = -x +c$$ $$ y^{-2} =2x +c$$ $$y(0)=0 \implies c=y_0^{-2}$$
$$ y= \frac {1}{\sqrt {2x+c}} = \frac {1}{\sqrt {2x+y_0^{-2}}} = \frac {y_0}{\sqrt {2xy_0^2 +1}} $$
The interval of existence is as far as you don't cross the vertical asymptote of $$\frac {y_0}{\sqrt {2xy_0^2 +1}} $$ which depends on the initial value.