The degree of the differential equation satisfied by the curves $\sqrt{1 + x} - a\sqrt{1 + y} = 1$, is ____.
Step-by-Step Solution
Key Concept: Differentiating the parametric family of curves once eliminates the arbitrary constant $a$, yielding a first-order differential equation.
Given the family of curves $\sqrt{1+x} - a\sqrt{1+y} = 1$, we differentiate both sides with respect to $x$ to eliminate the parameter $a$. Differentiating: $\frac{1}{2\sqrt{1+x}} - a\cdot\frac{1}{2\sqrt{1+y}}\cdot\frac{dy}{dx} = 0$. This gives us $\frac{1}{2\sqrt{1+x}} = \frac{a}{2\sqrt{1+y}}\cdot\frac{dy}{dx}$. From the original equation, we can express $a = \frac{\sqrt{1+x}-1}{\sqrt{1+y}}$. Substituting this eliminates $a$ and gives us a differential equation involving only $x$, $y$, and $\frac{dy}{dx}$. The highest derivative in the resulting equation is $\frac{dy}{dx}$, which is of order 1.
Correct Answer: 1