Show that # y^2 = (4x)(a-x) = 4ax-4x^2 # is a solution to the DE? # 2xy dy/dx = y^2 - 4x^2#

As the quoted Differential Equation is malformed, let us work back from the quoted solution to form the correct DE:

Correction of the Question:

If a solution is:

# y^2 = (4x)(a-x) = 4ax-4x^2 #

is a solution, then implicit differentiation gives:

# \ \ \ \ \ 2y dy/dx = 4a-8x #

Multiplying by #x# gives us:

# \ \ \ \ \ 2xy dy/dx = 4ax-8x^2 #
# :. 2xy dy/dx = 4ax-4x^2 - 4x^2#
# :. 2xy dy/dx = y^2 - 4x^2#

Therefore the question should (presumably) be to solve the DE

# 2xy dy/dx = y^2 - 4x^2#

And # x=a/2 => y^2 = (4a/2)(a-a/2) = a^2 => y = a#

Solution:

We have:

# 2xy dy/dx = y^2 - 4x^2# ..... [A]

As suggested, let us perform the substitution:

# y = vx iff v = y/x #
# :. dy/dx = v(d/dx x) + (d/dx v )x #
# :. dy/dx = v + x(dv)/dx #

Substituting into the (corrected) DE [A] we have:

# 2x(vx) (v + x(dv)/dx) = (vx)^2 - 4x^2 #

# :. 2v^2x^2 + 2vx^3(dv)/dx = v^2x^2 - 4x^2 #

# :. 2v^2 + 2vx(dv)/dx = v^2 - 4 #

# :. 2vx(dv)/dx = -(v^2 + 4) #

# :. (2v)/(v^2 + 4) (dv)/dx = -1/x #

This is now a First Order separable DE, so we can "separate the variables" to get:

# int \ (2v)/(v^2 + 4) \ dv = - \ int \ 1/x \ dx #

And now we can integrate, to get:

# :. ln(v^2 + 4) = - ln x + C #

Applying the initial conditions:

# y = a# when #x = a/2 #
# y = vx => a = (va)/2 #
# :. v=2 #

So we have modified initial conditions #v=2# when #x=a/2#:

# ln(4+4) = - ln (a/2) + C => C = ln(a/2)+ln8 =ln 4a #

So our particular solution is:

# ln(v^2 + 4) = - ln x + ln(4a) #

# :. ln(v^2 + 4) = ln((4a)/x) #

# :. v^2 + 4 = (4a)/x #

Restoring the substitution:

# (y/x)^2 + 4 = (4a)/x #

# :. y^2/x^2 + 4 = (4a)/x #

# :. y^2 + 4x^2 = 4ax #

# :. y^2 = 4ax - 4x^2 #

# :. y^2 = 4x(a-x) \ \ \ # QED