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Solve `[2sqrt(x y)-x]dy+ydx=0` |
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Answer» Correct Answer - `log_(e)y+sqrt(x/y)=C` We have `(dy)/(dx)=(y/x)/(1-2sqrt(y/x))` which is homogeneous. Put `y=vx` so that `(dy)/(dx) = x(dv)/(dx)+v` `therefore x(dv)/(dx) = v/(1-2sqrt(v))-v=(2v^(3//2))/(1-2sqrt(v))` or `(dx)/x=(1-2sqrt(v))/(2v^(3//2))dv=1/(2v^(3//2)-1/v)dv` Integrating, we get `-C+logx=-v^(-1//2)-logv=-sqrt(x/y)-logy+logx` or `logy + sqrt(x/y)=C` |
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