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Order and Degree of y²(y″)²+3x(y′)^⅓+x²y²=sinx | JEE

JEE Maths question with a full step-by-step solution.

Question
If pp and qq are the order and degree of the differential equation y2(d2ydx2)2+3x(dydx)1/3+x2y2=sin⁡xy^2\left(\dfrac{d^2y}{dx^2}\right)^2 + 3x\left(\dfrac{dy}{dx}\right)^{1/3} + x^2y^2 = \sin x, then:
Ap>qp > q
Bpq=12\dfrac{p}{q} = \dfrac{1}{2}
Cp=qp = q
Dp<qp < qcorrect
Solution
The highest derivative present is d2ydx2\dfrac{d^2y}{dx^2}, so the order is p=2p = 2. For the degree, the equation must be made polynomial in the derivatives. Isolate the fractional power and cube:
3x(dydx)1/3=sin⁡x−x2y2−y2(d2ydx2)23x\left(\frac{dy}{dx}\right)^{1/3} = \sin x - x^2y^2 - y^2\left(\frac{d^2y}{dx^2}\right)^2
⇒27x3dydx=[sin⁡x−x2y2−y2(d2ydx2)2]3.\Rightarrow 27x^3\frac{dy}{dx} = \left[\sin x - x^2y^2 - y^2\left(\frac{d^2y}{dx^2}\right)^2\right]^3.
 On the right, (d2ydx2)2\left(\dfrac{d^2y}{dx^2}\right)^2 is cubed, giving (d2ydx2)6\left(\dfrac{d^2y}{dx^2}\right)^6. So the power of the highest-order derivative is q=6q = 6. ∴p=2<6=q\therefore p = 2 < 6 = q. Correct answer: (4)
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