Question

Find general solutions for the following equations. To earn full credit, you must show all work. (This includes showing all of the algebra required to solve these equations!) Box your final answers. a.) y′′ −25y=0 b.) y′′ +25y=0 c.) y′′ −10y′ +29y=0 d.) y′′′ −y′′ −4y′ −6y=0 e.) y⁽⁴⁾ +8y′′ +16y=0

Answer 1

Final Answers:

General solutions are:

a.)
`\( y(x) = c₁e^(5x) + c₂e^(-5x) \)`

b.)
`\( y(x) = c₁cos(5x) + c₂sin(5x) \)`

c.)
`\( y(x) = e^(5x)(c₁cos(2x) + c₂sin(2x)) \)`

d.)
`\( y(x) = c₁e^(2x) + c₂e^(-3x) + c₃e^(-2x) \)`

e.)
`\( y(x) = c₁e^(-2x) + c₂xe^(-2x) + c₃cos(2x) + c₄sin(2x) \)`

Step-by-step explanation:

a.) The given differential equation is
`\( y'' - 25y = 0 \)`. The characteristic equation is
`\( r² - 25 = 0 \)`, which factors into
`\( (r - 5)(r + 5) = 0 \).` The roots are
`\( r = 5 \) and \( r = -5 \)`, so the general solution is
`\( y(x) = c₁e^(5x) + c₂e^(-5x) \).`

b.) For
`\( y'' + 25y = 0 \`), the characteristic equation is
`\( r² + 25 = 0 \), leading to \( r = ±5i \)`. The general solution is
`\( y(x) = c₁cos(5x) + c₂sin(5x) \)`, using Euler's formula to express the trigonometric functions in terms of exponentials.

c.) Solving
`\( y'' - 10y' + 29y = 0 \)`gives a characteristic equation
`\( r² - 10r + 29 = 0 \)`. The roots are complex, leading to
`\( r = 5 ± 2i \)`. The general solution is
`\( y(x) = e^(5x)(c₁cos(2x) + c₂sin(2x)) \)`, combining exponential and trigonometric functions.

d.) The differential equation
`\( y''' - y'' - 4y' - 6y = 0 \)` has characteristic equation
`\( r³ - r² - 4r - 6 = 0 \)`, and its roots are
`\( r = 2, -3, -1 \)`. Thus, the general solution is
`\( y(x) = c₁e^(2x) + c₂e^(-3x) + c₃e^(-2x) \).`

e.) The differential equation
`\( y⁽⁴⁾ + 8y'' + 16y = 0 \)` can be solved by factoring
`\( (r² + 4)² = 0 \)`, leading to repeated roots
`\( r = ±2i \)`. The general solution is
`\( y(x) = c₁e^(-2x) + c₂xe^(-2x) + c₃cos(2x) + c₄sin(2x) \).`