# Solve the Differential Equation by using Substitution?

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(x)dx + (y-2x )dy = 0

it is homogeneous, so I know you multiply by either 1/x or 1/y, but I cannot really get beyond this. Please show as much work as possible…thanks

• x dx + (y – 2x)dy = 0

sub u = y/x

xu = y

u dx + x du = dy

x dx + (y – 2x)dy = 0

x dx + (xu – 2x)(u dx + x du) = 0

x dx + xu² dx + x²u du – 2xu dx – 2x² du = 0

(x + xu² – 2xu)dx + (x²u – 2x²) du = 0

x(1 + u² – 2u)dx + x²(u – 2) du = 0

(1 + u² – 2u)dx = x(2 – u) du

dx/x = (2 – u) du/(u² – 2u + 1)

dx/x = (2 – u) du/(u – 1)²

∫dx/x = ∫(2 – u) du/(u – 1)²

sub v = u – 1

dv = du

u = v + 1

∫dx/x = ∫(2 – (v + 1)) dv/v²

∫dx/x = ∫(1 – v) dv/v²

∫dx/x = ∫(1/v² – 1/v) dv

ln | x | = -1/v – ln|v| + C

Back substitute

ln | x | = -1/(u + 1) – ln|u + 1| + C

ln | x | = -1/(y/x + 1) – ln|y/x + 1| + C

simplify

ln | x | = -x/(y + x) – ln|(y + x)/x| + C

ln | x | = -x/(y + x) – ln|y + x| + ln|x| + C

0 = -x/(y + x) – ln|y + x| + C

x/(y + x) + ln|y + x| = C

Making either x or y the subject is difficult

• Substitution? I don’t know what you mean by 1/x or 1/y, but here’s how I would solve it. I’d use a change of variables. Is that what you mean?

There’s a theorem for homogeneous D.E.’s. If M(x,y)dx+N(x,y)dy=0 is homogeneous, then it can be solved by making the substitution y=vx.

xdx+(y-2x)dy=0 Use y=vx, and thus dy=xdv+vdx

xdx+(vx-2x)(xdv+vdx)=0

xdx+vx^2dv+v^2xdx-2x^2dv-2vxdx=0 (Check my algebra on that)

(x+v^2x-2vx)dx+(vx^2-2x^2)dv=0 (Move some stuff around…)

x(1+v^2-2v)dx=-x^2(v-2)dv (Divide by x)

(1+v^2-2v)dx=-x(v-2)dv (finally….)

-dx/x=(v-2)/(1+v^2-2v)dv (NOW we can integrate.)

-ln|x|=integral of (v-2)/(v^2-2v+1)dv

I stopped here because I think you can perform the integration. Once you get everything in terms of x and v, then from the first substitution, v=y/x. Then it’s just simplifying. I did this in just a couple minutes, so no guarantee I made no mistakes in the algebra. But the theorem is solid.

• Let x = vy. Then dx = vdy + ydv. So

vy(vdy + ydv) + (y – 2vy)dy = 0

v^2ydy + vy^2dv + (y – 2vy)dy = 0

vy^2dv + (y – 2vy + v^2y)dy = 0

vy^2dv + y(1 – 2v + v^2)dy = 0

vdv/(1 – 2v + v^2) + dy/y = 0

vdv/(v – 1)^2+ + dy/y = 0

(v – 1 + 1)dv/(v – 1)^2+ + dy/y = 0

dv/(v – 1) + dv/(v – 1)^2+ + dy/y = 0

ln|v – 1| – 1/(v – 1) + ln|y| = C

ln|x/y – 1| – 1/(x/y – 1) + ln|y| = C

ln|(x – y)/y| – 1/(x/y – 1) + ln|y| = C

ln|x – y| – ln|y| – y/(x – y) + ln|y| = C

ln|x – y| = C + y/(x – y)

x – y = Ke^[y/(x – y)]

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