differential equation 2

(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

5 Answers

62
Lokesh Verma ·

take 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

1
kartik sondhi ·

I am Getting (B)

1
rupal jain ·

my ans is ax2......a is a constant
but the ans to this ques is (c) any constant

62
Lokesh Verma ·

the 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 !

1
Surbhi Agrawal ·

oops!!!sorry.... i have corrected it!!..

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