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The degree of the differential equation satisfying the relation `sqrt(1+x^2) + sqrt(1+y^2) = lambda (x sqrt(1+y^2)- ysqrt(1+x^2))` isA. 1B. 2C. 3D. None of these |
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Answer» Correct Answer - a On putting x = tan A and y = tan B in the given relation , we get ` cos A + cos B = lambda ( sin A - sinB)` ` rArr tan.((A-B)/2) = 1/lambda ` ` rArr tan^(-1) x - tan^(-1) y = 2 tan^(-1) (1/lambda )` On differentiating w.r.t x, we get `1/(1+x^(2)) - 1/(1+y^(2)) = 0 ` ` rArr (dy)/(dx) = (1+y^(2))/(1+x^(2))` Clearly , it is a differential equation of degree ` |
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