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Concept 03 / 06Differential Equations

Homogeneous Differential Equations and the Substitution y = vx

Recognising homogeneous functions of degree zero, solving dy/dx = F(y/x) with y = vx, the alternative x = vy, and reducing (a1 x + b1 y + c1)/(a2 x + b2 y + c2) to homogeneous form by shifting the origin.

A function f(x,y)f(x, y) is homogeneous of degree nn if f(λx,λy)=λnf(x,y)f(\lambda x, \lambda y) = \lambda^{n}f(x, y) for all λ>0\lambda > 0. For example, x2+xyx^{2} + xy has degree 22, x2+y2\sqrt{x^{2} + y^{2}} has degree 11, and x−yx+y\frac{x - y}{x + y} has degree 00. A first-order DE is homogeneous if it can be written as dydx=F(yx)\frac{dy}{dx} = F\left(\frac{y}{x}\right), which happens exactly when dydx=P(x,y)Q(x,y)\frac{dy}{dx} = \frac{P(x, y)}{Q(x, y)} with PP and QQ homogeneous of the same degree. Quick test: replace xx by λx\lambda x and yy by λy\lambda y; if λ\lambda cancels completely from the right side, the equation is homogeneous. The substitution. Put y=vxy = vx, so dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}. The equation becomes v+xdvdx=F(v)v + x\frac{dv}{dx} = F(v), which separates: dvF(v)−v=dxx\frac{dv}{F(v) - v} = \frac{dx}{x}. Integrate, then replace vv by yx\frac{y}{x}. Geometrically, homogeneous equations have slope fields that are constant along each line through the origin (the slope depends only on the direction yx\frac{y}{x}), so the family of solutions is invariant under scaling from the origin. When xx appears more naturally as a function of yy, as in dxdy=G(xy)\frac{dx}{dy} = G\left(\frac{x}{y}\right), substitute x=vyx = vy instead. This is often simpler when the equation contains terms like ex/ye^{x/y}. If F(v)=vF(v) = v for some constant v0v_{0} (that is, F(v0)−v0=0F(v_{0}) - v_{0} = 0), the line y=v0xy = v_{0}x is itself a solution, which may be lost when dividing by F(v)−vF(v) - v. Reducible to homogeneous. For dydx=a1x+b1y+c1a2x+b2y+c2\frac{dy}{dx} = \frac{a_{1}x + b_{1}y + c_{1}}{a_{2}x + b_{2}y + c_{2}}: If a1a2≠b1b2\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}, the lines a1x+b1y+c1=0a_{1}x + b_{1}y + c_{1} = 0 and a2x+b2y+c2=0a_{2}x + b_{2}y + c_{2} = 0 meet at a point (h,k)(h, k). Shift the origin there: x=X+hx = X + h, y=Y+ky = Y + k. The constants disappear and the equation becomes homogeneous in X,YX, Y: dYdX=a1X+b1Ya2X+b2Y\frac{dY}{dX} = \frac{a_{1}X + b_{1}Y}{a_{2}X + b_{2}Y}. If a1a2=b1b2\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}, the lines are parallel, and the numerator and denominator contain the same combination a1x+b1ya_{1}x + b_{1}y (up to a factor). Put v=a1x+b1yv = a_{1}x + b_{1}y, which makes the equation separable. A special case: if b1=−a2b_{1} = -a_{2}, the equation can be rearranged as an exact differential, a1x dx+b1(y dx+x dy)+…a_{1}x\,dx + b_{1}(y\,dx + x\,dy) + \ldots, and integrated directly. Key Takeaways: - Homogeneous: dydx=F(yx)\frac{dy}{dx} = F(\frac{y}{x}); numerator and denominator of equal degree. - Substitute y=vxy = vx, y′=v+xv′y' = v + xv'; separate dvF(v)−v=dxx\frac{dv}{F(v) - v} = \frac{dx}{x}. - Use x=vyx = vy when the equation is naturally in xy\frac{x}{y}. - Linear fractional right side with intersecting lines: shift the origin to the intersection. - Parallel lines: substitute v=a1x+b1yv = a_{1}x + b_{1}y. - Lines y=v0xy = v_{0}x with F(v0)=v0F(v_{0}) = v_{0} are solutions too. - Pitfall: forgetting to replace vv by yx\frac{y}{x} at the end. Brain-Teaser 1: Solve (x2+y2)dx−2xy dy=0(x^{2} + y^{2})dx - 2xy\,dy = 0. Solution: dydx=x2+y22xy\frac{dy}{dx} = \frac{x^{2} + y^{2}}{2xy}, homogeneous of degree 00. Put y=vxy = vx: v+xv′=1+v22vv + xv' = \frac{1 + v^{2}}{2v}, so xv′=1−v22vxv' = \frac{1 - v^{2}}{2v} and 2v dv1−v2=dxx\frac{2v\,dv}{1 - v^{2}} = \frac{dx}{x}. Integrating: −ln⁡∣1−v2∣=ln⁡∣x∣+C-\ln|1 - v^{2}| = \ln|x| + C, so x(1−v2)=kx(1 - v^{2}) = k. Replacing v=yxv = \frac{y}{x}: x2−y2=kxx^{2} - y^{2} = kx. The solutions are hyperbolas (and the lines y=±xy = \pm x for k=0k = 0), all passing through the origin, which is typical of homogeneous equations. Brain-Teaser 2: Solve dydx=x+2y−32x+y−3\frac{dy}{dx} = \frac{x + 2y - 3}{2x + y - 3}. Solution: The lines x+2y=3x + 2y = 3 and 2x+y=32x + y = 3 meet at (1,1)(1, 1). Put x=X+1x = X + 1, y=Y+1y = Y + 1: dYdX=X+2Y2X+Y\frac{dY}{dX} = \frac{X + 2Y}{2X + Y}. With Y=vXY = vX: v+Xv′=1+2v2+vv + Xv' = \frac{1 + 2v}{2 + v}, so Xv′=1−v22+vXv' = \frac{1 - v^{2}}{2 + v}. Partial fractions: 2+v(1−v)(1+v)=3/21−v+1/21+v\frac{2 + v}{(1 - v)(1 + v)} = \frac{3/2}{1 - v} + \frac{1/2}{1 + v}. Integrating: −32ln⁡∣1−v∣+12ln⁡∣1+v∣=ln⁡∣X∣+C-\frac{3}{2}\ln|1 - v| + \frac{1}{2}\ln|1 + v| = \ln|X| + C. Multiplying by 22 and exponentiating: 1+v(1−v)3=kX2\frac{1 + v}{(1 - v)^{3}} = kX^{2}. With v=YXv = \frac{Y}{X} this becomes X+Y=k(X−Y)3X + Y = k(X - Y)^{3}, so the solution is x+y−2=k(x−y)3x + y - 2 = k(x - y)^{3}. Shifting the origin removed the constants and made the substitution possible. Brain-Teaser 3: Solve (1+ex/y)dx+ex/y(1−xy)dy=0\left(1 + e^{x/y}\right)dx + e^{x/y}\left(1 - \frac{x}{y}\right)dy = 0. Solution: The equation is naturally in xy\frac{x}{y}, so put x=vyx = vy, dx=v dy+y dvdx = v\,dy + y\,dv. Substituting: (1+ev)(v dy+y dv)+ev(1−v)dy=0(1 + e^{v})(v\,dy + y\,dv) + e^{v}(1 - v)dy = 0. Collect dydy: [v+vev+ev−vev]dy=(v+ev)dy[v + ve^{v} + e^{v} - ve^{v}]dy = (v + e^{v})dy. So (v+ev)dy+y(1+ev)dv=0(v + e^{v})dy + y(1 + e^{v})dv = 0, giving (1+ev)dvv+ev=−dyy\frac{(1 + e^{v})dv}{v + e^{v}} = -\frac{dy}{y}. Integrating: ln⁡∣v+ev∣=−ln⁡∣y∣+C\ln|v + e^{v}| = -\ln|y| + C, so y(v+ev)=ky(v + e^{v}) = k, that is, x+yex/y=kx + ye^{x/y} = k. Choosing x=vyx = vy rather than y=vxy = vx matched the structure of the exponent and kept the algebra short.

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Homogeneous Differential Equations and the Substitution y = vx — DPP

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