Question

For what values of r does the function y = `e^(rx)` satisfy the differential equation 5y'' + 14y' − 3y = 0?

a) For what values of r does the function y = `e^(rx)` satisfy the differential equation 5y'' + 14y' − 3y = 0? (Enter your answers as a comma-separated list.)

r = ???

(b) If `r_1` and `r_2` are the values of r that you found in part (a), show that every member of the family of functions y = `ae^(r_1x) + be^(r_2x)` is also a solution. (Let `r_1` be the larger value and `r_2` be the smaller value.)

Answer 1

(a) r = `1/5`, -3

(b) Shown in explanation.

Explanation:

(b) We replace y with `ae^(r_1 x) + be^(r_2 x)`. First, we must figure out the values for both y' and y''.

y = `ae^(x/5) + be^(-3x)`
y' = `1/5 ae^(x/5) - 3 be^(-3x)`
y'' = `1/25 ae^(x/5) - 9 be^(-3x)`

Now we can insert the values in our differential equation 5y'' + 14y' − 3y = 0.

`5[1/25 ae^(x/5) - 9 be^(-3x)] + 14[1/5 ae^(x/5) - 3 be^(-3x)] − 3[ae^(x/5) + be^(-3x)] = 0`

`a[1/5 e^(x/5) + 14/5 e^(x/5) - 3 e^(x/5)] + b[45 e^(-3x) - 42 e^(-3x) - 3 e^(-3x)] = 0`

`a[3 e^(x/5) - 3 e^(x/5)] + b[45 e^(-3x) - 42 e^(-3x) - 3 e^(-3x)] = 0`

`a[3 e^(x/5) - 3 e^(x/5)] + b[3 e^(-3x) - 3 e^(-3x)] = 0`

a[0] + b[0] = 0

0 = 0

Since the solutions are both 0, it means that every member of the family of functions in y = `ae^(r_1 x) + be^(r_2 x)` is a solution to 5y'' + 14y' − 3y = 0.

Answer 2

`r=1/5` or `-3`; For details please see below.

Explanation:

(a) As `y=e^(rx)`

`y'=(dy)/(dx)=re^(rx)`

and `y''=r^2e^(rx)`

Hence `5y'' + 14y' − 3y = 0` can be written as

`5r^2e^(rx)+14re^(rx)-3e^(rx)-0`

or `5r^2+14r-3=0`

or `5r^2+15r-r-3=0`

or `(5r-1)(r+3)=0`

i.e. `r=1/5` or `-3`

(b) Consider the two solutions as `r_1` and `r_2`, then we have `5r_1^2+14r_1-3=0` and `5r_2^2+14r_2-3=0`

As `y=ae^(r_1x)+be^(r_2x)`, then

`y'=ar_1e^(r_1x)+br_2e^(r_2x)`

and `y''=ar_1^2e^(r_1x)+br_2^2e^(r_2x)`

`:.` `5y''+14y'−3y=5ar_1^2e^(r_1x)+5br_2^2e^(r_2x)+14ar_1e^(r_1x)+14br_2e^(r_2x)-3ae^(r_1x)-3be^(r_2x)`

= `ae^(r_1x)(5r_1^2+14r_1-3)+be^(r_2x)(5r_2^2+14r_2-3)`

= `0`

Hence `y=ae^(r_1x)+be^(r_2x)` is also a solution of `5y''+14y'−3y=0`