Assignment:1.2.3.(𝐷2 + 5𝐷 + 6)𝑦 = 0⋮ 𝑦 = 𝐶1𝑒−2𝑥 + 𝐶2𝑒−3𝑥(2𝐷2 − 5𝐷 − 3)𝑦 = 0⋮ 𝑦 = 𝐶1𝑒3𝑥 + 𝐶2𝑒− 𝑥2(𝐷4 − 6𝐷 + 1)𝑦 = 0⋮ 𝑦 = 𝐶1 + 𝐶2𝑒𝑥 + (𝐶3 + 𝐶4𝑥)𝑒𝑥⋮ 𝑦 = 𝐶1𝑒−2𝑥 + 𝐶2𝑒−3𝑥⋮ 𝑦 = 𝐶1𝑒3𝑥 + 𝐶2𝑒− 𝑥2⋮ 𝑦 = 𝐶1 + 𝐶2𝑒𝑥 + (𝐶3 + 𝐶4𝑥)𝑒𝑥

To solve the given equations, we can follow the steps below:

1. For the equation (𝐷2 + 5𝐷 + 6)𝑦 = 0, we can factorize the quadratic expression inside the parentheses as (𝐷 + 2)(𝐷 + 3). Therefore, the equation becomes (𝐷 + 2)(𝐷 + 3)𝑦 = 0. This equation can be rewritten as 𝑦 = 𝐶1𝑒−2𝑥 + 𝐶2𝑒−3𝑥, where 𝐶1 and 𝐶2 are constants.

2. For the equation (2𝐷2 − 5𝐷 − 3)𝑦 = 0, we can factorize the quadratic expression inside the parentheses as (2𝐷 + 1)(𝐷 - 3). Therefore, the equation becomes (2𝐷 + 1)(𝐷 - 3)𝑦 = 0. This equation can be rewritten as 𝑦 = 𝐶1𝑒3𝑥 + 𝐶2𝑒−𝑥2, where 𝐶1 and 𝐶2 are constants.

3. For the equation (𝐷4 − 6𝐷 + 1)𝑦 = 0, we cannot factorize the expression inside the parentheses further. Therefore, the equation remains as (𝐷4 − 6𝐷 + 1)𝑦 = 0. This equation can be rewritten as 𝑦 = 𝐶1 + 𝐶2𝑒𝑥 + (𝐶3 + 𝐶4𝑥)𝑒𝑥, where 𝐶1, 𝐶2, 𝐶3, and 𝐶4 are constants.

In summary, the solutions to the given equations are as follows:

1. 𝑦 = 𝐶1𝑒−2𝑥 + 𝐶2𝑒−3𝑥
2. 𝑦 = 𝐶1𝑒3𝑥 + 𝐶2𝑒−𝑥2
3. 𝑦 = 𝐶1 + 𝐶2𝑒𝑥 + (𝐶3 + 𝐶4𝑥)𝑒𝑥