# Maths

Substituting y=vx to transform following Differential Equations into a Recognisable form
1)(2xy)(dy/dx) +x^2+y^2=0
2)(x^2-y^2)dy/dx=xy

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1. so, did you do that?
If y = vx, then
dy/dx = v + x * dv/dx

(2xy)(dy/dx) +x^2+y^2=0
(2x*vx)(v + x * v') + x^2 + (vx)^2 = 0
2x^3 vv' + 3x^2v^2 + x^2 = 0
2x^3 v dv = -(x^2 + 3x^2v^2) dx
v dv = -1/(2x) (1+3v^2) dx
v/(3v^2+1) dv = -1/(2x) dx
Now it's simple to arrive at
v = √(c-x^3)/(√3 x^(3/2))
so, y = vx = √(c-x^2)/√(3x)

See what you can do with the other one, ok?

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