# Thread: First Order Homogeneous Diff EQ

1. ## First Order Homogeneous Diff EQ

For some reason, this is giving me fits. Im supposed to put it in the form: $M(x,y)dx+N(x,y)dy=0$ then use the substitution $v=\frac{y}{x}$, so $y=vx$ and $dy=vdx+xdv$

Here is the D.E.

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

Is it safe to make it like this: $(x-y)dy=\sqrt{x^2+y^2}dx$ then $\sqrt{x^2+y^2}dx-(x-y)dy=0$ then use the substitution?

Thanks

2. if you multiply by dx you get:

xdy = ydx + sqrt(x^2 +y^2)dx

(y + sqrt(x^2+y^2))dx -xdy = 0

y =vx

using dy = vdx + xdv => (vx + sqrt(x^2+v^2x^2)dx - x(vdx+xdv) = 0

(vx + x sqrt(1+v^2) -xv)dx - x^2dv = 0

(xsqrt(1+v^2))dx -x^2 dv = 0

which now separates 1/x dx = dv/sqrt(1+v^2)

Typically this is not how you would approach this problem--no need to put

in the form Mdx + Ndy = 0.

dy/dx = y/x + sqrt[1+(y/x)^2]

simply y = vx dy/dx = v + xdv/dx

v+ xdv/dx = v +sqrt(1+v^2)

dv/sqrt(1+v^2) = dx/x

By the way if dy/dx = f(y/x) we call it a differnetial equation with homogeneous coefficients which can be solved by letting y = vx

3. Ahh, I see..

I got $(x\sqrt{1+v^2})dx=x^2dv$ then to $\frac{(\sqrt{1+v^2})dx}{x}=dv$ then to bring the $\sqrt{1+v^2}$ over under the dv by diving, so then you get $\frac{dx}{x}=\frac{dv}{\sqrt{1+v^2}}$ then integrate, and solve...sweet!

Thanks for the help.