Solve:dy/dx=-(3x^2 y+y^2)/(2x^3+3xy
Let v=y/x or y=vx. Then dy/dx=v+x∙dv/dx
So then, v+x∙dv/dx=-(3x^2 y+y^2)/(2x^3+3xy)
x∙dv/dx=-(3x^2 y+y^2)/(2x^3+3xy)-(2xv+3v^2)/(2x+3v)=-(5xv+4v^2)/(2x+3v)
dv/dx=-(5xv+4v^2)/(2x^2+3xv)
dv/dx=-(5(x/v)+4(v/x)^2)/(2+3(v/x) )= -(5w+4w^2)/(2+3w), w=v/x
Let v=wx, then dv/dx=w+x∙dw/dx
So, w+x∙dw/dx=-(5w+4w^2)/(2+3w)
Then, x∙dw/dx=-(5w+4w^2)/(2+3w)-(2w+3w^2)/(2+3w)= -(7w+7w^2)/(2+3w)
The above DE now transforms to,
1/7 ∫(2+3w)/(w+w^2 ) dw=-∫x dx
Partial fractions
(2+3w)/(w+w^2 )=A/w+B/(1+w)
From which we find, A=2,B=1
Then,
1/7 ∫(2+3w)/(w+w^2 ) dw=1/7 {∫2/w dw+∫1/(1+w) dw}=1/7 {2.ln(w)+ln(1+w) }
1/7 ∫(2+3w)/(w+w^2 ) dw=1/7∙ln(w^2+w^3 )
And,
-∫1/x dx= -ln(x)+C
Therefore,
1/7∙ln(w^2+w^3 )=C - ln(x)=ln(A/x) (ln(A) = C)
w^2+w^3=kx^(-7), (k = A^7)
Where w=v/x and v=y/x, giving w=y/x^2 .
So,
w^2+w^3=y^2/x^4 +y^3/x^6 =kx^(-7)
Or,
(y^2 x^3+y^3 x=k),k≠0
If we differentiate the above implicit equation and solve for y’, we will get the original DE in the problem statement.