If \( y=x\displaystyle\frac{dy}{dx}+\sqrt{a^{2}\left(\displaystyle\frac{dy}{dx}\right)^{2}+b^{2}} \) , then the order and degree of the differential equation are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → order = 1, degree = 2
The only derivative present is \( \frac{dy}{dx} \), so the order is 1.
The degree is read only after the equation is made free of radicals in the derivative. With the square root on one side and both sides squared:
\( \displaystyle y-x\frac{dy}{dx}=\sqrt{a^2\left(\frac{dy}{dx}\right)^2+b^2} \)
\( \displaystyle \left(y-x\frac{dy}{dx}\right)^2=a^2\left(\frac{dy}{dx}\right)^2+b^2 \)
\( \displaystyle (x^2-a^2)\left(\frac{dy}{dx}\right)^2-2xy\frac{dy}{dx}+y^2-b^2=0 \)
This is a polynomial in \( \frac{dy}{dx} \) with highest power 2, so the degree is 2.
Tip: The exponent 2 in \( \left(\frac{dy}{dx}\right)^2 \) is a power, not the order; the order counts the highest derivative, which here is only the first.