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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))` is |
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Answer» Correct Answer - `(dy)/(dx) = (1+y^(2))/(1+x^(2))` Putting x=tanA and y=tanB in the given relation, we get cosA + cosB=`lambda`(sinA-sinB) or `tan^(-1)x-tan^(-1)y=2tan^(-1)(1/lambda)` Differentiating w.r.t. to x, we get `1/(1+x^(2))-1/(1+y^(2))(dy)/(dx)=0` or `(dy)/(dx)=(1+y^(2))/(1+x^(2))` |
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