Question

The equation x2ydx-(x3+y3)dy= 0 is homogeneous equation

Answer 1

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$<b>Thanks$

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Answer 2

if we assume y/x = t 
y = xt 
dy/dx = t + xdt/dx 

then, t/(1 + t³) = t + x.dt/dx 

t/(1 + t³) - t = xdt/dx 

(t - t - t⁴)/(1 + t³) = x dt/dx 

-t⁴/(1 + t³) = x dt/dx 

dx/x = -(1 + t³)/t⁴ dt 

lnx = -t^-4 dt - dt/t 

lnx = -t^-3/-3 - lnt + C 

lnx = 1/3t³ - lnt + C 

now, put t = y/x 

so, lnx = 1/3(y/x)³ - lny/x + C 

or, lnx = x³/3y³ - lny + lnx + C 

hence, -x³/3y³ + lny + C = 0