Answer
(a)
The solution of this differential equation is $y=\dfrac{1}{x^2-x-6}$.
(c)
The interval in which the solution is defined is $R-\{3,-2\}$.
(b)
Work Step by Step
(a)
We have given $\dfrac{dy}{dx}=(1-2x)y^2$
Now solve the differential equation as follows:
$\dfrac{dy}{dx}=(1-2x)y^2$
$\dfrac{dy}{y^2}=(1-2x)dx$
Since, $\dfrac{1}{a}=a^{-1}$
We get, $y^{-2}\,dy=(1-2x)dx$
On simplifying
We get, $y^{-2}\,dy=dx-2xdx$
Now integrate both sides as follows:
$\int y^{-2}\,dy=\int dx-2\int xdx$
We get, $-y^{-1}=x-x^2+C$
Or, $y^{-1}=x^2-x+C_0$
Or, $\dfrac{1}{y}=x^2-x+C_0$
Thus the explicit form is $y=\dfrac{1}{x^2-x+C_0}$.
We have given the initial condition $y(0) = \dfrac{−1}{6}$.
Substitute $x=0$ and $y = \dfrac{−1}{6}$
We get, $\dfrac{-1}{6}=\dfrac{1}{0^2-0+C_0}$
$\implies C_0=-6$
Hence, the solution of this differential equation is $y=\dfrac{1}{x^2-x-6}$.
(c)
We know the denominator can not be zero for a rational function.
Therefore, $x^2-x-6\ne 0$
$\implies x^2-3x+2x-6\ne 0$
$\implies x(x-3)+2(x-3) \ne 0$
$\implies (x-3)(x+2)\ne0$
$\implies x\ne3$ and $x\ne-2$
Hence, the interval in which the solution is defined is $R-\{3,-2\}$.
(b)
