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Concept 01 / 06Differential Equations

Variable Separable Equations and Reducible Forms

Separating variables and integrating both sides, handling the constant of integration and initial conditions, the substitution v = ax + by + c for dy/dx = f(ax + by + c), and checking for lost solutions when dividing.

A first-order DE is separable if it can be written as dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y). Then move all yy terms to one side and all xx terms to the other: dyg(y)=f(x) dx\frac{dy}{g(y)} = f(x)\,dx, and integrate: ∫dyg(y)=∫f(x) dx+C\int\frac{dy}{g(y)} = \int f(x)\,dx + C. Only one constant is needed, since the constants from the two sides combine. The result is often an implicit relation between xx and yy; solve for yy if the question asks for it or if it is easy. Choosing the form of the constant. Since CC is arbitrary, it can be written in whatever form makes the answer neat: ln⁡∣y∣=x2+C\ln|y| = x^{2} + C becomes y=Aex2y = Ae^{x^{2}} with A=±eCA = \pm e^{C}; ln⁡∣y∣=ln⁡∣x∣+C\ln|y| = \ln|x| + C becomes y=kxy = kx. When you exponentiate, the constant usually becomes a multiplier, and the sign can be absorbed into it. Lost solutions. Dividing by g(y)g(y) assumes g(y)≠0g(y) \neq 0. If g(y0)=0g(y_{0}) = 0 for some constant y0y_{0}, then y=y0y = y_{0} is a constant solution that may not appear in the general formula. For dydx=y2\frac{dy}{dx} = y^{2}, separating gives −1y=x+C-\frac{1}{y} = x + C, but y=0y = 0 is also a solution. Such equilibrium solutions matter when an initial condition places you on them. Initial value problems. After finding the general solution, substitute the initial condition to determine CC. Doing this at the end is safest; doing it immediately after integrating (with definite integrals from the initial point) is equivalent and sometimes faster: ∫y0ydyg(y)=∫x0xf(x)dx\int_{y_{0}}^{y}\frac{dy}{g(y)} = \int_{x_{0}}^{x}f(x)dx. Reducible to separable. If dydx=f(ax+by+c)\frac{dy}{dx} = f(ax + by + c), substitute v=ax+by+cv = ax + by + c. Then dvdx=a+bdydx=a+bf(v)\frac{dv}{dx} = a + b\frac{dy}{dx} = a + bf(v), which is separable: dva+bf(v)=dx\frac{dv}{a + bf(v)} = dx. For example, dydx=(x+y)2\frac{dy}{dx} = (x + y)^{2} with v=x+yv = x + y gives dvdx=1+v2\frac{dv}{dx} = 1 + v^{2}, so tan⁡−1v=x+C\tan^{-1}v = x + C and y=tan⁡(x+C)−xy = \tan(x + C) - x. Useful separable forms involve products that factor, such as dydx=ex−y+x2e−y\frac{dy}{dx} = e^{x - y} + x^{2}e^{-y}: multiply through by eye^{y} to get eydy=(ex+x2)dxe^{y}dy = (e^{x} + x^{2})dx, so ey=ex+x33+Ce^{y} = e^{x} + \frac{x^{3}}{3} + C. Always look for a factorisation that splits the right side into (function of xx) times (function of yy). Key Takeaways: - dydx=f(x)g(y)⇒∫dyg(y)=∫f(x)dx+C\frac{dy}{dx} = f(x)g(y) \Rightarrow \int\frac{dy}{g(y)} = \int f(x)dx + C. - One constant suffices; reshape it (ln⁡C\ln C, eCe^{C}) for a neat answer. - Check for constant solutions where g(y)=0g(y) = 0. - dydx=f(ax+by+c)\frac{dy}{dx} = f(ax + by + c): put v=ax+by+cv = ax + by + c. - Apply initial conditions after integrating, or integrate between the initial and general points. - Pitfall: splitting dydx=f(x)+g(y)\frac{dy}{dx} = f(x) + g(y) as if it were separable; sums do not separate. Brain-Teaser 1: Solve dydx=1+y21+x2\frac{dy}{dx} = \frac{1 + y^{2}}{1 + x^{2}} with y(0)=1y(0) = 1. Solution: Separate: dy1+y2=dx1+x2\frac{dy}{1 + y^{2}} = \frac{dx}{1 + x^{2}}, so tan⁡−1y=tan⁡−1x+C\tan^{-1}y = \tan^{-1}x + C. At x=0x = 0, y=1y = 1: π4=C\frac{\pi}{4} = C. So tan⁡−1y−tan⁡−1x=π4\tan^{-1}y - \tan^{-1}x = \frac{\pi}{4}. Taking tangents, y−x1+xy=1\frac{y - x}{1 + xy} = 1, giving y=1+x1−xy = \frac{1 + x}{1 - x} for x<1x < 1. Converting an inverse trigonometric relation into an algebraic one with a tangent formula produces a clean explicit answer. Brain-Teaser 2: Solve dydx=sin⁡(x+y)+cos⁡(x+y)\frac{dy}{dx} = \sin(x + y) + \cos(x + y). Solution: Put v=x+yv = x + y: dvdx=1+sin⁡v+cos⁡v\frac{dv}{dx} = 1 + \sin v + \cos v. Using half-angle identities, 1+cos⁡v+sin⁡v=2cos⁡2v2+2sin⁡v2cos⁡v2=2cos⁡2v2(1+tan⁡v2)1 + \cos v + \sin v = 2\cos^{2}\frac{v}{2} + 2\sin\frac{v}{2}\cos\frac{v}{2} = 2\cos^{2}\frac{v}{2}(1 + \tan\frac{v}{2}). So ∫12sec⁡2v21+tan⁡v2dv=x+C\int\frac{\frac{1}{2}\sec^{2}\frac{v}{2}}{1 + \tan\frac{v}{2}}dv = x + C, which gives ln⁡∣1+tan⁡v2∣=x+C\ln|1 + \tan\frac{v}{2}| = x + C. The solution is ln⁡∣1+tan⁡x+y2∣=x+C\ln\left|1 + \tan\frac{x + y}{2}\right| = x + C. The substitution makes the equation separable; the half-angle identity makes the resulting integral standard. Brain-Teaser 3: A curve passes through (1,1)(1, 1), and at every point (x,y)(x, y) on it the slope of the tangent is twice the slope of the line joining the point to the origin. Find the curve. Solution: The line to the origin has slope yx\frac{y}{x}, so dydx=2yx\frac{dy}{dx} = \frac{2y}{x}, which separates to dyy=2dxx\frac{dy}{y} = \frac{2dx}{x}, so ln⁡∣y∣=2ln⁡∣x∣+C\ln|y| = 2\ln|x| + C and y=Ax2y = Ax^{2}. The point (1,1)(1, 1) gives A=1A = 1: the curve is y=x2y = x^{2}. Checking: the slope of y=x2y = x^{2} at (x,x2)(x, x^{2}) is 2x=2x2x2x = \frac{2x^{2}}{x}, as required. Geometric conditions on the tangent translate directly into first-order DEs.

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Variable Separable Equations and Reducible Forms — DPP

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