(3y^3 − 5yx^2)dx + (xy^2+x^3)dy = 0 ; x=1 , y=2

Question

HOMOGENEOUS DE

(3y^3 − 5yx^2) + (xy^2+x^3) = 0 ; x=1 , y=2

Answer 1

Let y=vx, then y'=v+xv'.

(3y³-5yx²)dx+(xy²+x³)dy=0,

(3y³-5yx²)+(xy²+x³)y'=0,

(3v³x³-5vx³)+(v²x³+x³)(v+xv')=0,

3v³-5v+(v²+1)(v+xv')=0,

3v³-5v+v³+xv²v'+v+xv'=0,

4v³-4v+xv'(v²+1)=0,

x(dv/dx)(v²+1)=4v-4v³,

(v²+1)dv/(v-v³)=4dx/x.

Partial fractions:

(v²+1)/(v(1-v²))=A/v+B/(1-v)+C/(1+v),

v²+1≡A-Av²+Bv+Bv²+Cv-Cv²;

Equating coefficients of like terms:

A=1; -A+B-C=1, B-C=2; B+C=0, C=-B⇒2B=2, B=1, C=-1.

∫(v²+1)dv/(v-v³)=4∫dx/x becomes:

∫dv/v+∫dv/(1-v)-∫dv/(1+v)=4ln|ax| where a is a constant,

ln|v|-ln|1-v|-ln|1+v|=4ln|ax|=ln(a⁴x⁴)=ln(cx⁴) where c=a⁴ a positive constant.

v/(1-v²)=cx⁴, (y/x)/(1-y²/x²)=cx⁴, equating log arguments,

xy/(x²-y²)=cx⁴, y/(x²-y²)=cx³.

This can be written various ways: y=cx³(x²-y²)=cx⁵-cx³y², cx³y²+y-cx⁵=0, ...

x=1, y=2 satisfies the equation:

2=-3c, c=-⅔, 3y+2x³(x²-y²)=0 or 2x⁵-2x³y²+3y=0.