Question

If a and b are respectively the order and degree of a 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:

\textbf{Order:} The order of the highest derivative in the DE.
\textbf{Degree:} The highest power of the highest order derivative, after the DE has been cleared of radicals and fractional powers with respect to the derivatives.

Answer 1

The Correct Option is C

The given differential equation is \( y^2(y'')^2 + 3x(y')^{1/3} + x^2y^2 = \sin x \). To determine the degree, we must first clear fractional powers of derivatives. Isolate the term with the fractional power: \( 3x(y')^{1/3} = \sin x - x^2y^2 - y^2(y'')^2 \) Cube both sides to eliminate the cube root: \( (3x(y')^{1/3})^3 = (\sin x - x^2y^2 - y^2(y'')^2)^3 \) \( 27x^3 (y')^{ (1/3) \times 3 } = (\sin x - x^2y^2 - y^2(y'')^2)^3 \) \( 27x^3 y' = (\sin x - x^2y^2 - y^2(y'')^2)^3 \) Now the equation is a polynomial in derivatives. Order (a): The order is the order of the highest derivative present. The highest derivative is \(y''\) (second derivative). So, order \(a = 2\). Degree (b): The degree is the highest power of the highest order derivative after the equation is made polynomial in derivatives. The highest order derivative is \(y''\). In the term \( (\sin x - x^2y^2 - y^2(y'')^2)^3 \), when expanded, the term involving the highest power of \(y''\) will come from \( (-y^2(y'')^2)^3 \). \( (-y^2(y'')^2)^3 = (-1)^3 (y^2)^3 ((y'')^2)^3 = -y^6 (y'')^6 \). The power of the highest derivative \(y''\) is 6. So, degree \(b = 6\). We have \(a=2\) and \(b=6\). Check the options: % Option (a) b=a \(\implies 6=2\) (False) % Option (b) a=3b \(\implies 2=3(6) \implies 2=18\) (False) % Option (c) b=3a \(\implies 6=3(2) \implies 6=6\) (TRUE) % Option (d) ab=6 \(\implies (2)(6)=12 \neq 6\) (False) Thus, the correct relation is \(b=3a\). \[ \boxed{b=3a} \]