Solving First-order Linear ODEs | Georgia Southern University

University:
Georgia Southern University
Course:
MATH 3230 | Ordinary Differential Equations
Academic year:

2024
Views:

86

Pages:

10
Author:

Jalen W.

Definition A first-order ODE is said to be linear if it can be put in the following form for some single variable (univariate) functions p(x) and q(x). dy + p(x) · y = q(x) dx If it does not have the above form then it is nonlinear. Example Specify whether the following ODEs are linear or nonlinear. I dy dx = 2y + 3x I I I I dy y −x dx = x−1 dy dx + sin(xy ) = 1 dy y dx = 3x dy 3 dx = 2y + 3x 3x Solving First-order Linear ODEs We are going to create something called an integrating factor, which is just a special function (say, µ(x)) that we will multiply our ODE by (hence, ‘factor’) in order to perform a necessary integration (hence, ‘integrating’). We define µ(x) in the following way. Z R p(x) dx ] [ p(x) dx µ(x) = e = exp Note that we also have the following. d [R p(x) dµ = [e dx dx dx ] ] = e[ R p(x) dx ] d · dx Z p(x) dx = µ(x) · p(x) Solving First-order Linear ODEs Now, multiply both sides of our first-order linear ODE by µ(x). dy + p(x) · y dx dy µ(x) · + p(x) · y dx dy µ(x) · + µ(x) · p(x) · y dx dy dµ µ(x) · + ·y dx dx d [µ(x) · y (x)] dx = q(x) = µ(x) · q(x) = µ(x) · q(x) = µ(x) · q(x) = µ(x)q(x) Solving First-order Linear ODEs Now, we can integrate to solve the ODE. Z Z d [µ(x) · y (x)] dx = µ(x)q(x) dx dx Z µ(x) · y (x) = µ(x)q(x) dx Z 1 µ(x)q(x) dx y (x) = µ(x) Thus, we have our solution, y (x), provided we can perform the integration. Example Solve the following ODE. dy + 3y = x 2 dx Example Solve the following ODE. dy + 3y = sin(2x) dx Example Solve the following IVP. dy + 3xy = 3x, y (0) = 2 dx Example Solve the following IVP. dy 1 1 + y= , y (0) = 1 dx x +1 (x + 1)2 Example Solve the following IVP. dy 1 1 − y= , y (0) = 1 dx x +1 (x + 1)2

Solving First-order Linear ODEs

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