solution of the equation `(dy)/(dx)+1/x tany=1/x^2 tanysiny` isA. ` 2x= (1-2Cx^(2))sin y`B. `x=(1-2 Cx^(2)) sin y`C. `2x = (1+2Cx^(2))sin y`D. None of these

solution of the equation `(dy)/(dx)+1/x tany=1/x^2 tanysiny` isA. ` 2x

1.	solution of the equation `(dy)/(dx)+1/x tany=1/x^2 tanysiny` isA. ` 2x= (1-2Cx^(2))sin y`B. `x=(1-2 Cx^(2)) sin y`C. `2x = (1+2Cx^(2))sin y`D. None of these
Answer» Correct Answer - d The given differential equation can be rewritten as ` cot y "cosec" y (dy)/(dx) +1/x "cosec" y = 1/(x^(2))` On putting ` v = - "cosec" y and (dv)/(dx) = "cosec" y cot y (dy)/(dx),` we get ` (dv)/(dx) -1/x v = 1/(x^(2))` which is linear differential equation in v. Here, `P = -1/x and Q = 1/(x^(2))` ` :. IF = e^(int Pdx)= e^(int - 1/xdx)= e^(-logx)=1/x` Required solution is ` v = 1/x = int 1/x * 1/(x^(2))dx +C = -1/(2x^(2))+C` ` rArr 2x = (1- 2Cx^(2) ) sin y `

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solution of the equation `(dy)/(dx)+1/x tany=1/x^2 tanysiny` isA. ` 2x= (1-2Cx^(2))sin y`B. `x=(1-2 Cx^(2)) sin y`C. `2x = (1+2Cx^(2))sin y`D. None of these

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