If differential equation of first degree of a curve is given by $x^2\left(\frac{dy}{dx}\right)^2 - x(2y - 1)\frac{dy}{dx} + (y^2 - y - 2) = 0$, then $(y - 2015 . x)$ is a positive prime number '$P$' then the value of $P$ is ____.
Step-by-Step Solution
Key Concept: The quadratic differential equation factors into two linear first-order differential equations, whose solutions are straight lines of the form $y = mx + c$.
The given differential equation is $x^2\left(\frac{dy}{dx}\right)^2 - x(2y-1)\frac{dy}{dx} + (y^2-y-2) = 0$. This is a quadratic in $\frac{dy}{dx}$. Factoring: $x^2\left(\frac{dy}{dx}\right)^2 - x(2y-1)\frac{dy}{dx} + (y-2)(y+1) = 0$ gives $(x\frac{dy}{dx} - (y-2))(x\frac{dy}{dx} - (y+1)) = 0$. This yields two cases: $\frac{dy}{dx} = \frac{y-2}{x}$ or $\frac{dy}{dx} = \frac{y+1}{x}$. Solving the first: $\frac{dy}{y-2} = \frac{dx}{x}$ gives $y - 2 = Cx$, so $y - 2 = Cx$ is one family of solutions. For the second case, we get $y + 1 = Dx$. The expression $y - 2015x$ becomes a constant when the curve follows $y = 2015x + c$ for some constant $c$. For this to be a positive prime number, we need $P = 2$, which is the only even prime.
Correct Answer: 2