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The general solution of the differential equation ` (dy)/(dx) = ((1+y^(2)))/(xy(1+x^(2)))` isA. `(1+x^(2))(1+y^(2))=C`B. `(1+x^(2))(1+y^(2))=Cx^(2)`C. `(1-x^(2))(1-y^(2))=C`D. `(1+x^(2))(1+y^(2))=Cy^(2)` |
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Answer» Correct Answer - b Given differential equation can be rewritten as ` y/(1+y^(2))dy= (dx)/(x(1+x^(2)))` `rArr 1/2 int (2y)/(1+y^(2))dy = (dx)/(x(1+x^(2)))` ` rArr 1/2 int (2y)/((1+y^(2)))dy=1/2 int (dt)/(t(t+1))` [ put `x^(2)` = t and 2x dx = dt in RHS integral ] ` rArr 1/2 int ((2y" "dy))/(1+y^(2))=1/2 int (1/t - 1/(1+t))dt` ` rArr 1/2 * log (1+y^(2)) = 1/2 [ log t - log (1+t)] +1/2 log C` ` rArr log 1+y^(2) = logx^(2)- log (1+x^(2))+log C` ` rArr log (1+y^(2))(1+x^(2))= log Cx^(2)` ` rArr (1+y^(2))(1+x^(2))= Cx^(2)` |
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