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Find the equation of a curve passing through `(0,1)`and having gradient `(1(y+y^3))/(1+x+x y)a t(x , y)` |
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Answer» Correct Answer - `xy+tan^(-1)y=pi/4` We have `(dy)/(dx)=(-y+y^(3))/(1+x(1+y^(2))` or `(dx)/(dy) = -(1+x(1+y^(3)))/(1+x(1+y^(2))` `=-1/(y(y^(2)+1))+x(1+y^(2))/(y(1+y^(2))]` `therefore (dx)/(dy) + x/y=-1/(y(1+y^(2))` I.F. `=e^(int(dy)/y)=e^(logy)=y`. Hence, the solution is `x.y=-int1/(y(1+y^(2)).ydy+C` `=-int(dy)/(1+y^(2))+C=-tan^(-1)y+C` or `xy+tan^(-1)y=C`. This passes through (0,1). Therefore, `tan^(-1)1=C, i.e., C=pi/4`. Thus, the equation of the curve is `xy+tan^(-1)y=pi/4` |
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