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Equation of curve passing through `(1,1)` & satisfyng the differential equation `(xy^(2)+y^(2)+x^(2)y+2xy)dx+(2xy+x^(2))dy=0` isA. `xy(x+y)=2e^(1-x)`B. `xy(x^(2)+y^(2))=2e^(x-1)`C. `x^(2)y^(2)(x+y)=2`D. `x^(2)y^(2)(x^(2)+y^(2))=2` |
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Answer» Correct Answer - A `[(xy^(2)dx+(y^(2)dx+2xydy)]+x^(2)ydx+(2xydx+x^(2)dy)=0` `e^(x) xy^(2)dx+e^(x)(y^(2)dx+xd(y^(2)))+e^(x)x^(2)ydx+e^(x)(2xydx)+x^(2)dy)=0` `d(e^(x)xy^(2))+d(e^(x)x^(2)y)=0` `e^(x)xy^(2)+e^(x)x^(2)y=C` Passing through `(1,1)` so `C=2e` so, curve is `xy(x+y)=2e^(1-x) xy(x+y)=2e^(1-x)` |
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