(x-y) (4x + y)dx + (5x - y)dy =0

Question

homogenous equation

Answer 1

Let y=vx, dy/dx=v+xdv/dx.

The DE becomes:

(x-vx)(4x+vx)+(5x-vx)dy/dx=0,

(1-v)(4+v)+(5-v)(v+xdv/dx)=0,

4-3v-v²+5v+5xdv/dx-v²-vxdv/dx=0, 4+2v-2v²+5xdv/dx-vxdv/dx=0,

x(dv/dx)(5-v)=2v²-2v-4=2(v²-v-2).

This can be written:

[(5-v)/(v²-v-2)]dv=2dx/x and the whole expression is now integrable.

(5-v)/(v²-v-2) can be split into partial fractions:

(5-v)/[(v+1)(v-2)]≡A/(v+1)+B/(v-2),

5-v≡A(v-2)+B(v+1)=Av-2A+Bv+B, so: A+B=-1, -2A+B=5, 3A=-6, A=-2, B=1:

∫[-2/(v+1)+1/(v-2)]dv=2∫dx/x=ln(ax²), where a is a constant.

-ln((v+1)²)+ln|v-2|=ln(ax²), (v-2)/(v+1)²=ax², ((y/x)-2)/((y/x)+1)²=ax²,

[(y-2x)/x][x²/(x²+y²)]=ax², (y-2x)/(x²+y²)=ax.