Answer
(a)
The solution of this differential equation is $y=-\sqrt{2x-2x^2+4}$.
(c)
The interval in which the solution is defined is $[-1,2]$.
(b)
Work Step by Step
(a)
We have given $\dfrac{dy}{dx}=\dfrac{(1-2x)}{y}$
Now solve the differential equation as follows:
$\dfrac{dy}{dx}=\dfrac{(1-2x)}{y}$
$\implies y\,dy=(1-2x)dx$
On simplifying
We get, $y\,dy=dx-2x\,dx$
Now integrate both sides as follows:
$\int y\,dy=\int dx-2\int xdx$
We get, $\dfrac{y^2}{2}=x-x^2+C$
Or, $y^2=2x-2x^2+C_0$
Thus the explicit form can be $y=\sqrt{2x-2x^2+C_0}$ or $y=-\sqrt{2x-2x^2+C_0}$.
We have given the initial condition $y(1) = -2$.
So, $y=-\sqrt{2x-2x^2+C_0}$ is the only solution, since $y$ can be negative only in this solution.
Substitute $x=1$ and $y = -2$
We get, $-2=-\sqrt{2(1)-2(1)^2+C_0}$
$\implies C_0=4$
Hence, the solution of this differential equation is $y=-\sqrt{2x-2x^2+4}$.
(c)
We know that we cannot have a negative value inside a square root.
Therefore, $2x-2x^2+4\ge 0$
$\implies -x^2+x+2\ge 0$
$\implies -x^2+2x-x+2\ge 0$
$\implies x(-x+2)+1(-x+2) \ge 0$
$\implies (x+1)(2-x)\ge0$
$\implies 2\ge x$ and $x\ge-1$
Hence, the interval in which the solution is defined is $[-1,2]$.
(b)
