Consider the following differential equation. 6y'' + 3y' − 3y = 0. For what values of r does the function y = e^rx satisfy the equation?

Question

I have no idea how to approach these types of equations. Also anything with e variables I find highly intimidating. What are the first few steps to solving this problem?

Answer 1

It is the method of Undetermined Coefficients.

y = exp(rx)

y' = r(exp(rx)) <--- chain rule

y'' = r^2 * exp(r(x)

plugs these into the diff.eq

6 * r^2 exp(rx) + 3 * r * exp(rx) - 3 * exp(rx) = 0

(3exp(rx)) (2r^2 + r - 1) = 0

exponential function is always positive, never zero.

2r^2 + r - 1 = 0

( 2r - 1 )(r + 1 ) = 0

2r-1 = 0 ---> r = 1/2

r+1 = 0 --> r = -1

the two solutions are: exp( x/2) and exp(-x) = 1/exp(x)

You may have been taught about the

characteristic equation is 6r^2 + 3r - 3 = 0 where y'' maps to r^2, maps to y' and r maps to y

2r^2 + r - 1 = 0

checking:

y = exp(x/2) ---> y' = (1/2)exp(x/2) ---> y'' = (1/4)exp(x/2)

plug-in: 6(1/4) exp(x/2) + 3(1/2) exp(x/2) - 3 exp(x/2) = 0

y = exp(-x) ---> y' = -exp(-x) ---> y'' = exp(-x)

6 exp(-x) + -3 exp(-x) - 3 exp(-x) = 0