Question

Sum of the squares of the order and degree (if defined) of a differential equation $2y' + (y'')^2 = \sqrt{y'' - 3}$ is

• A. 13
• B. 20
• C. 8
• D. 16

Hint:

Never determine the degree of a differential equation while derivatives are under radicals, fractional powers, or inside transcendental functions (like sin, log, $e$). You must first algebraically manipulate the equation into a polynomial form of its derivatives.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given differential equation is: \[ 2y' + (y'')^2 = \sqrt{y'' - 3} \] We need to find:

• Order: Highest order derivative present
• Degree: Power of highest order derivative after removing radicals/fractions

Step 2: Find the Order:
The derivatives present are $y'$ and $y''$.
Highest order derivative is $y''$.
\[ \text{Order} = 2 \]
Step 3: Remove Radical to Find Degree:
Square both sides to eliminate the square root: \[ (2y' + (y'')^2)^2 = y'' - 3 \] Expand LHS: \[ (2y')^2 + 2(2y')(y'')^2 + ((y'')^2)^2 = y'' - 3 \] \[ 4(y')^2 + 4y'(y'')^2 + (y'')^4 = y'' - 3 \] Rearrange: \[ (y'')^4 + 4y'(y'')^2 - y'' + 4(y')^2 + 3 = 0 \]
Step 4: Find the Degree:
Now the equation is polynomial in derivatives.
Highest power of highest order derivative $y''$ is: \[ \text{Degree} = 4 \]
Step 5: Required Sum:
\[ \text{Sum} = (\text{Order})^2 + (\text{Degree})^2 \] \[ = 2^2 + 4^2 = 4 + 16 = 20 \]
Step 6: Final Answer:
\[ \boxed{20} \]