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.