Differential EquationsmediumFree

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

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

Question

p

and

q

are the order and degree of the differential equation

y^2\left(\dfrac{d^2y}{dx^2}\right)^2 + 3x\left(\dfrac{dy}{dx}\right)^{1/3} + x^2y^2 = \sin x

, then:

p > q

\dfrac{p}{q} = \dfrac{1}{2}

p = q

p < q

correct

Solution

The highest derivative present is

\dfrac{d^2y}{dx^2}

, so the order is

p = 2

. For the degree, the equation must be made polynomial in the derivatives. Isolate the fractional power and cube:

3x\left(\frac{dy}{dx}\right)^{1/3} = \sin x - x^2y^2 - y^2\left(\frac{d^2y}{dx^2}\right)^2

\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,

\left(\dfrac{d^2y}{dx^2}\right)^2

is cubed, giving

\left(\dfrac{d^2y}{dx^2}\right)^6

. So the power of the highest-order derivative is

q = 6

\therefore p = 2 < 6 = q

. Correct answer: (4)

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