I have the Cauchy problem: $y'+(y^2+y-3/4)cosx=0$ with $y(0)=y_o$. I have to found the maximal intervall of definition for the solution. If $y_0=1/2$and $y_0=-3/2$ the solution are constant and defined on all R. In the other cases can I to calculate the expression of the solution (the ED is a separable variables)?
I have found: $|\frac{2y(x)-1}{2y(x)+3}*\frac{2y_o+3}{2y_o-1}|=e^{-2sinx}$.
So $y(x)=-\frac{3}{2}+\frac{2y_0+3}{2y_0+3-(2y_0-1)e^{-2sinx}}$.
the intervall of definition is ${2y_0+3-(2y_0-1)e^{-2sinx}}\ne0$?
Just a hint... $$y'+(y^2+y-3/4)\cos(x)=0$$ $$y'+(y^2+y+\frac 14 -1)\cos(x)=0$$ $$y'+((y+\frac 12)^2 -1)\cos(x)=0$$ Substitute $z=y+\frac 12$ $$z'+z^2\cos(x)=\cos(x)$$ its separabe $$z'=(1-z^2)\cos(x)$$ $$\int \frac {dz}{(1-z^2)}=\int \cos(x)dx=\sin(x)+K$$ $$\ln|{1+z}|+\ln|{1-z}|=2\sin(x)+K$$ $$\ln|{\frac 32+y}|+\ln|{\frac 12-y}|=2\sin(x)+K$$ $$K=\ln|{\frac 32+y_0}|+\ln|{\frac 12-y_0}|$$ $$....$$