Differential-Equations Homogeneous Equations with Constant Coefficients, guaranteed and verified 100% Pass

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Homogeneous Equations with Constant Coefficients

To solve an 𝑛th order homogeneous linear differential equation with constant
coefficients:

𝑎𝑛 𝑦 (𝑛) + 𝑎𝑛−1 𝑦 (𝑛−1) + ⋯ + 𝑎1 𝑦 ′ + 𝑎0 𝑦 = 0
𝑎0 , 𝑎1 , … , 𝑎𝑛 ∈ ℝ
We guess at a solution: 𝑦 = 𝑒 𝑟𝑥 .

Given that the derivatives of 𝑦 as

𝑦′ = 𝑟𝑒 𝑟𝑥
𝑦′′ = 𝑟 2 𝑒 𝑟𝑥
⋮
𝑦 (𝑛) = 𝑟 𝑛 𝑒 𝑟𝑥
we get:

𝑎𝑛 𝑟 𝑛 𝑒 𝑟𝑥 + ⋯ + 𝑎1 𝑟𝑒 𝑟𝑥 + 𝑎0 𝑒 𝑟𝑥 = 0
(𝑎𝑛 𝑟 𝑛 + ⋯ + 𝑎1 𝑟 + 𝑎0 )𝑒 𝑟𝑥 = 0.
So we must solve the characteristic equation:

𝑎𝑛 𝑟 𝑛 + 𝑎𝑛−1 𝑟 𝑛−1 + ⋯ + 𝑎1 𝑟 + 𝑎0 = 0.


Theorem: If the roots of the characteristic equation are distinct real
numbers, 𝑟1 , … , 𝑟𝑛 , then we can say:

𝑦(𝑥 ) = 𝑐1 𝑒 𝑟1 𝑥 + 𝑐2 𝑒 𝑟2 𝑥 + ⋯ + 𝑐𝑛 𝑒 𝑟𝑛 𝑥
is a general solution to:

𝑎𝑛 𝑦 (𝑛) + 𝑎𝑛−1 𝑦 (𝑛−1) + ⋯ + 𝑎1 𝑦 ′ + 𝑎0 𝑦 = 0.

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Ex. Solve the initial value problem 𝑦 ′′′ + 2𝑦 ′′ − 3𝑦 ′ = 0 where

𝑦(0) = 2, 𝑦 ′ (0) = −7, 𝑦 ′′ (0) = 5.


The characteristic equation is:

𝑟 3 + 2𝑟 2 − 3𝑟 = 0
𝑟(𝑟 2 + 2𝑟 − 3) = 0
𝑟(𝑟 + 3)(𝑟 − 1) = 0
𝑟 = 0, −3, 1.


So the general solution is:

𝑦 = 𝑐1 𝑒 0 + 𝑐2 𝑒 (−3𝑥) + 𝑐3 𝑒 𝑥 = 𝑐1 + 𝑐2 𝑒 (−3𝑥) + 𝑐3 𝑒 𝑥
𝑦 ′ = −3𝑐2 𝑒 (−3𝑥) + 𝑐3 𝑒 𝑥
𝑦 ′′ = 9𝑐2 𝑒 (−3𝑥) + 𝑐3 𝑒 𝑥
2 = 𝑦(0) = 𝑐1 + 𝑐2 𝑒 0 + 𝑐3 𝑒 0 = 𝑐1 + 𝑐2 + 𝑐3
−7 = 𝑦 ′ (0) = −3𝑐2 + 𝑐3
5 = 𝑦 ′′ (0) = 9𝑐2 + 𝑐3


−7 = −3𝑐2 + 𝑐3
5 = 9𝑐2 + 𝑐3
−12 = −12𝑐2
⟹ 𝑐2 = 1, 𝑐3 = −4, 𝑐1 = 5.
So the solution to the initial value problem is:

𝑦 = 5 + 𝑒 (−3𝑥) − 4𝑒 𝑥 .