(y3-2x2y)dx + (2xy2 - x3)dy =0, then the value of xy√(y2-x2) is
1. y2 +x
2. xy2
3. any constant
4. none of these
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5 Answers
62take y = tx
you will get
\\-\frac{t^3-2t}{2t^2-1}=t+xdt/dx \\-\frac{t^3-2t}{2t^2-1}-t=xdt/dx \\-\frac{3t^3-3t}{2t^2-1}=xdt/dx \\-dx/x=\frac{2t^2-1}{3t^3-3t}dt
now solve it :)
I hope i have not made any calculation error :P
1my ans is ax2......a is a constant
but the ans to this ques is (c) any constant
62the question is given that =
the RHS is not clear.. can you make sure that the RHS is zero?
THat is why i din try it whole heartedly either !
