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Concept 06 / 06Differential Equations

Exact Differentials, Integrable Combinations and Orthogonal Trajectories

Recognising exact differentials such as d(xy), d(y/x), d(x^2 + y^2) and d(tan^-1(y/x)), the exactness test for M dx + N dy = 0, and finding the orthogonal trajectories of a family of curves by replacing dy/dx with -dx/dy.

Figure 1
Some DEs are best solved not by a general method but by spotting that one side is the differential of a known function. The standard integrable combinations: x dy+y dx=d(xy)x\,dy + y\,dx = d(xy). x dy−y dxx2=d(yx)\frac{x\,dy - y\,dx}{x^{2}} = d\left(\frac{y}{x}\right) and y dx−x dyy2=d(xy)\frac{y\,dx - x\,dy}{y^{2}} = d\left(\frac{x}{y}\right). x dy−y dxxy=d(ln⁡∣yx∣)\frac{x\,dy - y\,dx}{xy} = d\left(\ln\left|\frac{y}{x}\right|\right). x dy−y dxx2+y2=d(tan⁡−1yx)\frac{x\,dy - y\,dx}{x^{2} + y^{2}} = d\left(\tan^{-1}\frac{y}{x}\right). x dx+y dy=12d(x2+y2)x\,dx + y\,dy = \frac{1}{2}d(x^{2} + y^{2}). x dx+y dyx2+y2=d(x2+y2)\frac{x\,dx + y\,dy}{\sqrt{x^{2} + y^{2}}} = d\left(\sqrt{x^{2} + y^{2}}\right). x dy+y dxxy=d(ln⁡∣xy∣)\frac{x\,dy + y\,dx}{xy} = d(\ln|xy|) and x dy+y dxx2y2=d(−1xy)\frac{x\,dy + y\,dx}{x^{2}y^{2}} = d\left(-\frac{1}{xy}\right). exdy−yexdxe2x=d(yex)\frac{e^{x}dy - ye^{x}dx}{e^{2x}} = d\left(\frac{y}{e^{x}}\right). The skill is to group terms of the equation so that each group matches one of these, possibly after dividing the whole equation by a suitable factor such as x2x^{2}, xyxy or x2+y2x^{2} + y^{2}. The appearance of the pair x dy−y dxx\,dy - y\,dx is the typical signal. Exact equations. An equation M(x,y)dx+N(x,y)dy=0M(x, y)dx + N(x, y)dy = 0 is exact if the left side is the total differential dϕd\phi of some function ϕ(x,y)\phi(x, y); then the solution is ϕ(x,y)=C\phi(x, y) = C. The test is ∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} (partial derivatives: differentiate with respect to one variable, treating the other as a constant). To find ϕ\phi, integrate MM with respect to xx (treating yy as constant), then add the terms of NN that do not contain xx, integrated with respect to yy. For example, (2xy+3)dx+(x2+4y)dy=0(2xy + 3)dx + (x^{2} + 4y)dy = 0 has My=2x=NxM_{y} = 2x = N_{x}, so it is exact, and ϕ=x2y+3x+2y2\phi = x^{2}y + 3x + 2y^{2}. Orthogonal trajectories. Two families of curves are orthogonal trajectories of each other if every curve of one family meets every curve of the other at right angles. The figure shows the rectangular hyperbolas xy=cxy = c and y2−x2=ky^{2} - x^{2} = k. In physics, lines of force and equipotentials, and streamlines and velocity-potential curves, are orthogonal families. To find the orthogonal trajectories of a family: 1. Form the DE of the given family by eliminating its constant: F(x,y,y′)=0F(x, y, y') = 0. 2. Replace dydx\frac{dy}{dx} by −dxdy-\frac{dx}{dy} (slopes of perpendicular lines multiply to −1-1). 3. Solve the new DE. For xy=cxy = c: y+xy′=0y + xy' = 0 gives y′=−yxy' = -\frac{y}{x}. Replacing gives −dxdy=−yx-\frac{dx}{dy} = -\frac{y}{x}, so x dx=y dyx\,dx = y\,dy and y2−x2=ky^{2} - x^{2} = k. In polar coordinates the corresponding replacement is rdθdr→−1rdrdθr\frac{d\theta}{dr} \to -\frac{1}{r}\frac{dr}{d\theta}. It is essential to eliminate the constant before replacing the slope; otherwise the new equation still refers to the particular curve and the result is wrong. Key Takeaways: - x dy+y dx=d(xy)x\,dy + y\,dx = d(xy); x dy−y dxx2=d(yx)\frac{x\,dy - y\,dx}{x^{2}} = d(\frac{y}{x}); x dy−y dxx2+y2=d(tan⁡−1yx)\frac{x\,dy - y\,dx}{x^{2} + y^{2}} = d(\tan^{-1}\frac{y}{x}). - x dx+y dy=12d(x2+y2)x\,dx + y\,dy = \frac{1}{2}d(x^{2} + y^{2}). - Exact: My=NxM_{y} = N_{x}; solution ϕ=C\phi = C. - Orthogonal trajectories: DE of the family, then y′→−1y′y' \to -\frac{1}{y'}, then solve. - Eliminate the family's constant before replacing the slope. - Pitfall: grouping terms that only look similar, such as x dy+y dxx\,dy + y\,dx versus x dy−y dxx\,dy - y\,dx. Brain-Teaser 1: Solve x dy−y dx=(x2+y2)dxx\,dy - y\,dx = (x^{2} + y^{2})dx. Solution: Divide by x2+y2x^{2} + y^{2}: x dy−y dxx2+y2=dx\frac{x\,dy - y\,dx}{x^{2} + y^{2}} = dx. The left side is d(tan⁡−1yx)d\left(\tan^{-1}\frac{y}{x}\right). So tan⁡−1yx=x+C\tan^{-1}\frac{y}{x} = x + C, or y=xtan⁡(x+C)y = x\tan(x + C). Dividing by the right expression turned the equation into two exact differentials. (It is also homogeneous-like, but this route is shorter.) Brain-Teaser 2: Solve y dx−x dy+3x2y2ex3dx=0y\,dx - x\,dy + 3x^{2}y^{2}e^{x^{3}}dx = 0. Solution: Divide by y2y^{2}: y dx−x dyy2+3x2ex3dx=0\frac{y\,dx - x\,dy}{y^{2}} + 3x^{2}e^{x^{3}}dx = 0. The first part is d(xy)d\left(\frac{x}{y}\right) and the second is d(ex3)d(e^{x^{3}}). So xy+ex3=C\frac{x}{y} + e^{x^{3}} = C. The factor y2y^{2} was chosen because it turns y dx−x dyy\,dx - x\,dy into a known differential while also removing y2y^{2} from the other term. Brain-Teaser 3: Find the orthogonal trajectories of the family of parabolas y=ax2y = ax^{2}. Solution: Differentiate: y′=2axy' = 2ax. Eliminate a=yx2a = \frac{y}{x^{2}}: y′=2yxy' = \frac{2y}{x}. Replace y′y' by −dxdy-\frac{dx}{dy}: −dxdy=2yx-\frac{dx}{dy} = \frac{2y}{x}, so x dx=−2y dyx\,dx = -2y\,dy. Integrating: x22=−y2+C\frac{x^{2}}{2} = -y^{2} + C, that is, x2+2y2=kx^{2} + 2y^{2} = k. The orthogonal trajectories are ellipses centred at the origin. Each parabola through the origin crosses each ellipse at right angles.

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