Question

The differential equation whose solution is \( Ax^2 + By^2 = 1 \), where A and B are arbitrary constants is of}

• A. degree 1 and order 2
• B. degree 2 and order 1
• C. degree 3 and order 2
• D. degree 1 and order 3

Hint:

Order = Number of arbitrary constants. Degree is the power of the highest order derivative.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The number of arbitrary constants in a general solution indicates the order of the differential equation.
Step 2: Detailed Explanation:
There are two arbitrary constants, A and B. Therefore, the order of the differential equation must be 2.
Let's find the equation:
Differentiate once: \( 2Ax + 2Byy' = 0 \implies Ax + Byy' = 0 \dots (i) \)
Differentiate again: \( A + B(yy'' + (y')^2) = 0 \dots (ii) \)
From (i), \( A = -\frac{Byy'}{x} \). Substitute into (ii):
\[ -\frac{Byy'}{x} + B(yy'' + (y')^2) = 0 \] Since \( B \ne 0 \), divide by B:
\[ -\frac{yy'}{x} + yy'' + (y')^2 = 0 \implies xyy'' + x(y')^2 - yy' = 0 \] The highest order derivative is \( y'' \) (order 2) and its power is 1 (degree 1).
Step 4: Final Answer:
The differential equation has order 2 and degree 1.