Question

2. Find the solution of each differential equation. (a) y/2y-8y = 0 (b) 25y/- 20y + 4y = 0 (c) y + 2y + 2y = 0 2. Find the solution of each differential equation . ( a ) y / 2y - 8y = 0 ( b ) 25y / - 20y + 4y = 0 ( c ) y + 2y + 2y = 0​

Answer 1

Each of these ODEs is linear and homogeneous with constant coefficients, so we only need to find the roots to their respective characteristic equations.

(a) The characteristic equation for

`y'' - 2y' - 8y = 0`

is

`r^2 - 2r - 8 = (r - 4) (r + 2) = 0`

which arises from the ansatz
`y = e^(rx)`.

The characteristic roots are
`r=4` and
`r=-2`. Then the general solution is

`\boxed{y = C_1 e^(4x) + C_2 e^(-2x)}`

where
`C_1,C_2` are arbitrary constants.

(b) The characteristic equation here is

`25r^2 - 20r + 4 = (5r - 2)^2 = 0`

with a root at
`r=\frac25` of multiplicity 2. Then the general solution is

`\boxed{y = C_1 e^(2/5\,x) + C_2 x e^(2/5\,x)}`

(c) The characteristic equation is

`r^2 + 2r + 2 = (r + 1)^2 + 1 = 0`

with roots at
`r = -1 \pm i`, where
`i=√(-1)`. Then the general solution is

`y = C_1 e^((-1+i)x) + C_2 e^((-1-i)x)`

Recall Euler's identity,

`e^(ix) = \cos(x) + i \sin(x)`

Then we can rewrite the solution as

`y = C_1 e^(-x) (\cos(x) + i \sin(x)) + C_2 e^(-x) (\cos(x) - i \sin(x))`

or even more simply as

`\boxed{y = C_1 e^(-x) \cos(x) + C_2 e^(-x) \sin(x)}`