Question

Find a homogeneous linear differential equation with constant coefficients whose general solution is given by

Answer 1

Answer:

$y" + 2y' + 2y = 0$

Step-by-step explanation:

Given

$y=c_1e^{-x}cosx+c_2e^{-x}sinx$

Required

Determine a homogeneous linear differential equation

Rewrite the expression as:

$y=c_1e^{\alpha x}cos(\beta x)+c_2e^{\alpha x}sin(\beta x)$

Where

$\alpha = -1$ and $\beta = 1$

For a homogeneous linear differential equation, the repeated value m is given as:

$m = \alpha \± \beta i$

Substitute values for $\alpha$ and $\beta$

$m = -1 \± 1*i$

$m = -1 \± i$

Add 1 to both sides

$m +1= 1 -1 \± i$

$m +1= \± i$

Square both sides

$(m +1)^2= (\± i)^2$

$m^2 + m + m + 1 = i^2$

$m^2 + 2m + 1 = i^2$

In complex numbers:

$i^2 = -1$

So, the expression becomes:

$m^2 + 2m + 1 = -1$

Add 1 to both sides

$m^2 + 2m + 1 +1= -1+1$

$m^2 + 2m + 2= 0$

This corresponds to the homogeneous linear differential equation

$y" + 2y' + 2y = 0$