If p and q are order and degree of differential equation y 2 ( d x 2 d 2 y ) 2 + 3 x ( d x d y ) 3 1 + x 2 y 2 = sin x , then

y 2 ( d x 2 d 2 y ) 2 + x 2 y 2 − sin x = − 3 x ( d x d y ) 3 1 ( y 2 ( d x 2 d 2 y ) 2 + x 2 y 2 − sin x ) 3 = − 9 x 3 ( d x d y ) Here order = 2 = p Degree = 6 = q ∴ p < q

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MathematicsDifferential EquationGeneral (Order/degree/formation)Easy2 minQB

MathematicsEasysingle choice

If $p$ and $q$ are order and degree of differential equation $y^{2} (\frac{d ^{2} y}{d x ^{2}})^{2} + 3 x (\frac{d y}{d x})^{\frac{1}{3}} + x^{2} y^{2} = sin x$ , then

Options:

Answer:

Solution:

$y^{2} (\frac{d ^{2} y}{d x ^{2}})^{2} + x^{2} y^{2} - sin x = - 3 x (\frac{d y}{d x})^{\frac{1}{3}}$

$(y^{2} (\frac{d ^{2} y}{d x ^{2}})^{2} + x^{2} y^{2} - sin x)^{3} = - 9 x^{3} (\frac{d y}{d x})$

Here order $= 2 = p$

Degree $= 6 = q$

$∴ p < q$

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:General (Order/degree/formation)

⏱ 2mℹ️ Source: QB

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