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The solution of differential equation `(xy^(5)+2y)dx-xdy =0,` isA. `9 x^(8) + 4x^(9) y^(4) = 9y^(4)C`B. ` 9x^(8) - 4x^(9) y^(4) - 9y^(4) C = 0 `C. ` x^(8) (9+4y^(4)) = 10y^(4)C`D. None of these |
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Answer» Correct Answer - a `(xy^(5)+2y)dx = xdy` ` rArr x (dy)/(dx) - 2y = xy^(5)` ` rArr (dy)/(dx) -(2y)/x = y^(5)` ` rArr y^(-5) (dy)/(dx) - (2y^(-4))/x = 1" "` …(i) Put , `y^(-4) =t` ` rArr -4y^(-5) (dy)/(dx) = (-1)/4 (dt)/(dx)" "` …(ii) From Eqs. (i) and (ii) , ` -1/4 (dt)/(dx) - (2t)/x = 1 rArr (dt)/(dx) + (8t)/x =-4` ` IF = e^(int 8/xdx) = e^(8logx) = x^(8)` ` :. t* x^(8) int (-4)x^(8) dx +C` ` rArr (x^(8))/(y^(4))= -(4*x^(9))/9 +C` ` rArr 9x^(8) + 4x^(9)* y^(4) = 9y^(4)C` |
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