Question

If \(a\) and \(b\) are respectively the order and degree of the differential equation \[ y^2(y'')^2+3x(y')^{1/3}+x^2y^2=\sin x, \] then

• A. \(b=a\)
• B. \(a=3b\)
• C. \(b=3a\)
• D. \(ab=6\)

Hint:

The order is the highest derivative present. The degree is the power of the highest order derivative after the equation is made polynomial in derivatives.

Answer 1

The Correct Option is C

Step 1: Identify the order.
The highest derivative in \(y^2(y'')^2+3x(y')^{1/3}+x^2y^2=0\) is \(y''\), so order \(a=2\).

Step 2: Eliminate the fractional power to find the degree.
Isolate the fractional term: \(3x(y')^{1/3} = -y^2(y'')^2-x^2y^2\). Cube both sides: \(27x^3 y' = \left[-y^2(y'')^2-x^2y^2\right]^3\). The highest power of \(y''\) on the right is \((y'')^{2\times 3}=(y'')^6\). So degree \(b=6\).

Step 3: Check the relation.
\(b = 6 = 3\times 2 = 3a\). So \(b=3a\).
\[ \boxed{b = 3a} \]