Question

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

(portions of this question have been edited or deleted!)

Answer 1

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