Question

solve the following homogeneous equations

1. dy/dx=(2x-7y)/(3y-8x)
2. (2xy+3y²)dx-(2xy+x²)dy=0
3. xsin(y/x)(ydx+xdy)+ycos(y/x)(xdy-ydx)=0

Answer 1

Solution

1.$\frac{dy}{dx}=\frac{2x-7y}{3y-8x}$

Take

y=vx

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

v+x$\frac{dv}{dx}=\frac{2x-7vx}{8vx-8x}$

Divide the R.H.S with x and seperate by variables

$\frac{8v-8}{v-8v^2+2}dv=\frac1xdx$

Integrate both sides respectively,we have;

$\frac{-13ln(8v^2-v-2)+3\sqrt{65}(ln(16v+\sqrt{65}-1)-ln(16-\sqrt{65}-1)}{26}=ln(x)+C$

But v=$\frac yx$ ,replace in the equation;

$\frac{-ln(8\frac{y^2}{x^2}-\frac yx-2)+3\sqrt{65}(ln(16\frac yx+\sqrt{65}-1)-ln(16\frac yx-\sqrt{65}-1)}{26}=ln(x)+C$

2.

(2xy+3y²)dx-(2xy+x²)dy=0

Rewrite as follows,

$\frac{2xy+3y^2}{2xy+x^2}=\frac{dy}{dx}$

Take y=vx

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Replace

$\frac{2vx^2+3v^2x^2}{2vx^2+x^2}=v+x\frac{dv}{dx}$

Divide by x² and put like terms together.

$\frac{2v+1}{v^2+v}dv=\frac 1xdx$

Integrate both sides;

ln(v²+v)=ln(x)+C

Replace back v=$\frac yx$

ln($\frac{y^2}{x^2}$ +$\frac yx$)=ln(x)+C

3.

Rewrite as follows;

(chain(y/X)-y²cos(y/x)dx+(x²sin(y/x)-xycos(y/x)dy=0

$\frac{dy}{dx}$ =$\frac{y^2cos(\frac yx)-xysin(\frac yx)}{x^2sin(\frac yx)+xycos(\frac yx)}$

Take

y=vx and $\frac {dy}{dx}=v+x\frac {dv}{dx}$

Replace in the equation and divide x²

$v+x\frac{dv}{dx}=\frac{v^2cosv-vsinv}{sin v+vcosv}$

Seperate variables;

$\frac{sinv +vcosv}{-2vsinv}dv=\frac 1xdx$

Integrate both sides using suitable methods,we have;

$\frac{-ln(sin v)-ln( v)}{2}=ln(x)+C$

Replace back v=$\frac yx$

$\frac{-ln(sin(\frac yx)-ln(\frac yx)}{2}=ln(x )+C$