    Question

# If the order and degree of the differential equation satisfying √1−x2+√1−y2=b(x−y), where b is a parameter, is λ and μ respectively, then λ+μ is

A
4
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B
2
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C
3
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D
5
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Solution

## The correct option is B 2√1−x2+√1−y2=b(x−y) ⋯(1) Clearly, the order is one as there is only one independent parameter b. Put x=sinα, y=sinβ in eqn. (1) ⇒α=sin−1x, β=sin−1y Equation (1) becomes cosα+cosβ=b(sinα−sinβ) ⇒2cos(α+β2)cos(α−β2)=2bcos(α+β2)sin(α−β2) ⇒cot(α−β2)=b ⇒α−β=2cot−1b ⇒sin−1x−sin−1y=2cot−1b Differentiating w.r.t. x, we get 1√1−x2−1√1−y2dydx=0 Degree of above differential equation is one. ∴λ=μ=1 ⇒λ+μ=2  Suggest Corrections  0      Similar questions
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