the general solution of the equation y" - 8y' + 16y = 0 is
y = ye^4x - 2y₀xe^4x.
The general solution of the second order linear equation y" - 8y' + 16y = 0 is
y = C₁e^4x + C₂e²x,
where C₁ and C₂ are arbitrary constants.
We can find this solution by assuming that the solution is of the form y = e^(rx). Substituting this into the equation, we get
r²e^(rx) - 8re^(rx) + 16e^(rx) = 0.
Dividing both sides by e^(rx), we get
r² - 8r + 16 = 0.
Factoring this equation, we get
(r - 4)² = 0.
Therefore, r = 4 is a double root of the equation.
This means that the general solution of the equation is of the form y = (C₁ + C₂x)e^4x, where C₁ and C₂ are arbitrary constants.
To find the constants C₁ and C₂, we can use the fact that the solution must also satisfy the initial conditions y(0) = y₀ and y'(0) = y₁.
Substituting y = (C₁ + C₂x)e^4x into y(0) = y₀, we get
y₀ = (C₁ + 0)e^4·0 = C₁.
Substituting y = (C₁ + C₂x)e^4x into y'(0) = y₁, we get
y₁ = (4C₁ + C₂)e^4·0 = 4C₁.
Solving for C₁, we get C₁ = y₀. Substituting this into y = (C₁ + C₂x)e^4x, we get
y = (y₀ + C₂x)e^4x.
To find C₂, we can differentiate this equation, to find y'.
y' = (C₂ + 4y₀x)e^4x.
Substituting y = (y₀ + C₂x)e^4x and y' = (C₂ + 4y₀x)e^4x into y'' = 8y' - 16y, we get
e^4x(y₀ + C₂x)'' = 8e^4x(C₂ + 4y₀x) - 16e^4x(y₀ + C₂x).
Simplifying this equation, we get
C₂ = -2y₀.