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If x=g(y)ef(y) is a solution of the differential equation (1+y2)+(x−e−tan−1y)dydx=0, satisfying the condition x=0 at y=0, then the value of f(√2)+g(√2) is

Answer» If x=g(y)ef(y) is a solution of the differential equation (1+y2)+(xetan1y)dydx=0, satisfying the condition x=0 at y=0, then the value of f(2)+g(2) is


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