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7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables

Syllabus
2020
Topic
7.7
Level
—

Learning objectives

—FUN-7.E—Determine particular solutions to differential equations• FUN-7.E Determine particular solutions to differential equations.• FUN-7.E.1 A general solution may describe infinitely many solutions to a differential equation. There is only one particular solution passing through a given point.• FUN-7.E.2 The function F defined by F(x) = y₀ + ∫ₐˣ f(t)dt is a particular solution to the differential equation dy/dx = f(x), satisfying F(a) = y₀.• FUN-7.E.3 Solutions to differential equations may be subject to domain restrictions.• Enduring understanding FUN-7: Solving differential equations allows us to determine functions and develop models.

An Initial Condition Selects One Solution

A general solution contains an arbitrary constant and represents a family of functions. An initial condition such as y(a)=y0y(a)=y_0 determines that constant and selects the particular solution passing through (a,y0)(a,y_0).

F(x)=y_0+\int_a^x f(t),dt

When dy/dx=f(x)dy/dx=f(x), this formula automatically satisfies both requirements: the Fundamental Theorem of Calculus gives F′(x)=f(x)F'(x)=f(x), and F(a)=y0F(a)=y_0 because an integral with equal bounds is 00. The integration variable tt is a placeholder, leaving xx as the input of FF.

For dy/dx=y2dy/dx=y^2 with y(0)=1y(0)=1, separation gives y−2dy=dxy^{-2}dy=dx, so −1/y=x+C-1/y=x+C. Substituting (0,1)(0,1) gives C=−1C=-1, hence y=1/(1−x)y=1/(1-x). Differentiating yields y′=1/(1−x)2=y2y'=1/(1-x)^2=y^2, and y(0)=1y(0)=1, so both the equation and initial condition hold.

A particular formula is not automatically valid for every real xx. The solution 1/(1−x)1/(1-x) is undefined at x=1x=1, so the maximal interval containing the initial input 00 is (−∞,1)(-\infty,1). State a domain interval that contains the initial point and does not cross a discontinuity of the solution or differential equation.