Can any one solve: this diffrential eqn y= 2x(dy/dx) + (dy/dx)^2?

Question

maths:-diffrential eqn

Answer 1

put,
dy/dx = p


Hence,
y = 2px + p^2
or, p^2 + 2px - y = 0
or, p = [-2x (+/-)(4x^2 + 4y)^(1/2)]/2
or, p = [-x (+/-)(x^2 + y)^(1/2)]
or, dy/dx = [-x (+/-)(x^2 + y)^(1/2)]

Integrate to solve.
I am only showing the solution of, dy/dx= -x+(x^2+y)^(1/2)......(1)

put, x^2 + y = Q^2

Hence, 2x + dy/dx = 2Q*dQ/dx,
From (1) we have,
2x + dy/dx= x+(x^2+y)^(1/2)
or, 2Q * dQ/dx = x + p.............(2)

Now put, Q = vx,
hence dQ/dx = v + x * dv/dx,
Substituting in (2)
2vx * (v + x * dv/dx) = x + vx
or, 2v * (v + x * dv/dx) = 1 + v
or, 2v^2 + 2vx dv/dx = 1 + v
or, 2vx dv/dx = 1 + v - 2v^2
or, 2vdv/(1 + v - 2v^2) = dx/x
or, ∫2vdv/(1 + v - 2v^2) = ∫dx/x
or, ∫2vdv/(1 - v)(1 + 2v) = ∫dx/x
or, 2/3 *∫[(1 + 2v) - (1 -v)]dv/(1 - v)(1 + 2v) = ∫dx/x
or, 2/3 *[∫dv/(1 - v) - ∫dv/(1 + 2v)] = ∫dx/x
or, 2/3 [ - ln|1-v| - 1/2 ln|1 + 2v|] = ln|x| + c
or, 2/3 [ - ln|1-Q/x| - 1/2 ln|1 + 2Q/x|] = ln|x| + c
or, 2/3 [ - ln|1-(x^2 + y)½/x| - 1/2 ln|1 + 2(x^2 + y)½/x|] = ln|x| + c

(x^2 + y)½ means (x^2 + y)^½
Phewww............

Answer 2

initially substitute p=dy/dx and diff the eqn and i think u can make out

Answer 3

ANSWER IS 4^2
i dont no the meaning of ^ if it is miltiply than it would be 8 and if division than 2 and anything else than plz let me no

Answer 4

y= 2x(dy/dx) + (dy/dx)^2
or, y+x^2 = (x + dy/dx)^2
or √(y+x^2) = x + dy/dx
or, dy/dx = √(y+x^2) - x

integrate to solve

Answer 5

-000.563 *dt