Answer
(a)
The solution of this differential equation is $ y=\dfrac{-1+\sqrt{4x^2-15}}{2}$.
(c)
The interval in which the solution is defined is $\left(-\infty,-\sqrt{\dfrac{15}{4}}\right]\cup\left[\sqrt{\dfrac{15}{4}},\infty\right)$.
(b)
Work Step by Step
(a)
We have given $\dfrac{dy}{dx}=\dfrac{2x}{1+2y}$
Now solve the differential equation as follows:
$\dfrac{dy}{dx}=\dfrac{2x}{1+2y}$
$1dy+2y\,dy=2xdx$
Now integrate both sides as follows:
$\int1dy+\int2y\,dy=\int2xdx$
We get, $y^2+y=x^2+C$
Or, $y^2+y-x^2-C=0$
Use quadratic formula
We get, $ y=\dfrac{-1\pm\sqrt{1+4x^2+4C}}{2}$
Thus the explicit form can be $ y=\dfrac{-1+\sqrt{1+4x^2+4C}}{2}$ or $ y=\dfrac{-1-\sqrt{1+4x^2+4C}}{2}$.
We have given the initial condition $y(2) = 0$.
Substitute $x=2$ and $y =0$ in both equations.
We get, $ 0=\dfrac{-1-\sqrt{1+4(2)^2+4C}}{2}$
$\implies -1-\sqrt{1+4(2)^2+4C}=0$
$\implies -\sqrt{1+4(2)^2+4C}=1$
It is not possible since the square root of any real number is always positive. Thus, the left-hand side is negative which contradicts the fact that right-hand side is positive.
Now substitute the values in the second equation.
We get, $ 0=\dfrac{-1+\sqrt{1+4(2)^2+4C}}{2}$
$\implies -1+\sqrt{1+4(2)^2+4C}=0$
$\implies \sqrt{1+4(2)^2+4C}=1$
$\implies 1+4(2)^2+4C=1$
$\implies 16+4C=0$
$\implies C=-4$
Hence, the solution of this differential equation is $ y=\dfrac{-1+\sqrt{1+4x^2+4(-4)}}{2}$ or $ y=\dfrac{-1+\sqrt{4x^2-15}}{2}$.
(c)
We know that the value inside the square root can not be negative.
Therefore, $4x^2-15\ge 0$
$\implies 4x^2\ge 15$
$\implies x^2 \ge \dfrac{15}{4}$
$\implies x\ge\Big |\sqrt{\dfrac{15}{4}}\Big|$
$\implies x\le -\sqrt{\dfrac{15}{4}}$ and $x\ge \sqrt{\dfrac{15}{4}}$
Hence, the interval in which the solution is defined is $\left(-\infty,-\sqrt{\dfrac{15}{4}}\right]\cup\left[\sqrt{\dfrac{15}{4}},\infty\right)$.
(b)
