Q180 / 240 · KCET 2026 · Mathematics
Sum of the squares of the order and degree (if defined) of a differential equation $2 y^{\prime}+\left(y^{\prime \prime}\right)^2=\sqrt{y^{\prime \prime}-3}$ is
- A$3$
- Correct answerB$20$
- C$8$
- D$16$
Solution
Given differential equation is $2 y^{\prime} + (y^{\prime \prime})^2 = \sqrt{y^{\prime \prime} - 3}$
To find the degree, we must express the differential equation as a polynomial in its derivatives. Squaring both sides, we get:
$(2 y^{\prime} + (y^{\prime \prime})^2)^2 = y^{\prime \prime} - 3$
$4 (y^{\prime})^2 + 4 y^{\prime} (y^{\prime \prime})^2 + (y^{\prime \prime})^4 = y^{\prime \prime} - 3$
$(y^{\prime \prime})^4 + 4 y^{\prime} (y^{\prime \prime})^2 - y^{\prime \prime} + 4 (y^{\prime})^2 + 3 = 0$
The highest order derivative present in the equation is $y^{\prime \prime}$, so the order is $2$.
The highest power of the highest order derivative $y^{\prime \prime}$ is $4$, so the degree is $4$.
Sum of the squares of the order and degree is $2^2 + 4^2 = 4 + 16 = 20$.
Answer: $20$
To find the degree, we must express the differential equation as a polynomial in its derivatives. Squaring both sides, we get:
$(2 y^{\prime} + (y^{\prime \prime})^2)^2 = y^{\prime \prime} - 3$
$4 (y^{\prime})^2 + 4 y^{\prime} (y^{\prime \prime})^2 + (y^{\prime \prime})^4 = y^{\prime \prime} - 3$
$(y^{\prime \prime})^4 + 4 y^{\prime} (y^{\prime \prime})^2 - y^{\prime \prime} + 4 (y^{\prime})^2 + 3 = 0$
The highest order derivative present in the equation is $y^{\prime \prime}$, so the order is $2$.
The highest power of the highest order derivative $y^{\prime \prime}$ is $4$, so the degree is $4$.
Sum of the squares of the order and degree is $2^2 + 4^2 = 4 + 16 = 20$.
Answer: $20$
Hence, the correct answer is option: 'B'.