By substituting y=vx ; dy=vdy+ydv
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homogeneous differential equation solution for y(x²+y²)dx+x(3x²-5y²)dy=0

rewrite as,

dy/dx = -y(x^2 + y^2)/(x(3x^2 - 5y^2)) = -(y/x)(1 + (y/x)^2) / (3 - 5(y/x)^2)

let w = y/x. or y = wx, giving dy/dx = w + x.dw/dx

substituting for w = y/x and using dy/dx = w + x.dw/dx,

w + x.dw/dx = dy/dx = -w(1 + w^2) / (3 - 5w^2)

x.dw/dx = -w(1 + w^2) / (3 - 5w^2) - w = -w(1 + w^2) / (3 - 5w^2) - w(3 - 5w^2) / (3 - 5w^2)

x.dw/dx = (-w - w^3 - 3w + 5w^3) / (3 - 5w^2) = (-4w + 4w^3) / (3 - 5w^2) = -4w(1 - w^2) / (3 - 5w^2)

(3 - 5w^2) / (w(1 - w^2)) dw/dx = -4/x

integrating both sides wrt x,

int (3 - 5w^2) / (w(1 - w^2)) dw = int -4/x dx

Partial fractions on the lhs give us,

int {1/(w + 1) + 3/w + 1/(w - 1)} dw = int -4/x dx

ln(w+1) + ln(w^3) + ln(w-1) = ln(x^(-4)) + ln(K)

w^3(w^2 - 1) = Kx^(-4)

back-substituting for w = y/x,

(y/x)^3((y/x)^2 - 1) = Kx^(-4)

(y^3/x^5)(y^2 - x^2) = Kx^(-4)

Solution: y^3(y^2 - x^2) = Kx

 

by Level 11 User (81.5k points)
edited by

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