Homogeneous differential equations of first order

Proof. Let u = c 1 y 1 + c 2 y 2 . Then u = c 1 y 1 + c 2 y 2 and u = c 1 y 1 + c 2 y 2 . Since y 1 and y 2 are solutions of (1), ay 1 + by 1 + cy 1 = 0 and ay 2 + by 2 + cy 2 = 0. It follows that au + bu + cu = a(c 1 y 1 + c 2 y 2 ) + b(c 1 y 1 + c 2 y 2 ) + c(c 1 y 1 + c 2 y 2 ) = c 1 (ay 1 + by 1 + cy 1 ) + c 2 (ay 2 + by 2 + cy 2 ) = c 1 0 + c 2 0 = 0,

Chukwudi Mary Victory, 2021

This project work teasing and explaining to an extent, the method used in solving second order ordinary differential equation, with their application. More emphasis is laid on the use of variation of parameters and exact solutions for solution and transformation of higher order ordinary differential equation.

1. The rate of change in the population P (t) is the derivative P ′ (t). The

1. Second-order; linear.

1. Let y = e rt , so that y = r e rt and y = r 2 e rt. Direct substitution into the differential equation yields (r 2 + 3r − 4)e rt = 0. Canceling the exponential, the characteristic equation is r 2 + 3r − 4 = 0. The roots of the equation are r = −4 , 1. Hence the general solution is y = c 1 e t + c 2 e −4t. 2. Let y = e rt. Substitution of the assumed solution results in the characteristic equation r 2 + 5r + 6 = 0. The roots of the equation are r = −3 , −2. Hence the general solution is y = c 1 e −2t + c 2 e −3t. 4. Substitution of the assumed solution y = e rt results in the characteristic equation 3r 2 − 4r + 1 = 0. The roots of the equation are r = 1/3 , 1. Hence the general solution is y = c 1 e t/3 + c 2 e t. 6. The characteristic equation is 9r 2 − 16 = 0 , with roots r = ±4/3. Therefore the general solution is y = c 1 e −4t/3 + c 2 e 4t/3. 8. The characteristic equation is r 2 − 4r − 4 = 0 , with roots r = 2 ± 2 √ 2. Hence the general solution is y = c 1 e (2−2 √ 2)t + c 2 e (2+2 √ 2)t. 9. Substitution of the assumed solution y = e rt results in the characteristic equation r 2 + 2r − 3 = 0. The roots of the equation are r = −3 , 1. Hence the general solution is y = c 1 e −3t + c 2 e t. Its derivative is y = −3c 1 e −3t + c 2 e t. Based on the

muti, 2020

We will now turn our attention to nonhomogeneous second order linear equations, equations with the standard form y″ + p(t) y′ + q(t) y = g(t), g(t) ≠ 0. (*) Each such nonhomogeneous equation has a corresponding homogeneous equation: y″ + p(t) y′ + q(t) y = 0. (**) Note that the two equations have the same left-hand side, (**) is just the homogeneous version of (*), with g(t) = 0. We will focus our attention to the simpler topic of nonhomogeneous second order linear equations with constant coefficients: a y″ + b y′ + c y = g(t). Where a, b, and c are constants, a ≠ 0; and g(t) ≠ 0. It has a corresponding homogeneous equation a y″ + b y′ + c y = 0.