Find the complete general solution to the second order DE:
4y'' - 9y = 0
y = a e^kx
y' = a k e^kx
y" = a k^2 e^kx
4 a k^2 = 9 a
k = +/- 3/2
more interesting would be
4 y" + 9 y = 0
then you would get
k^2 = -9
and k = 1.5 i
so
y = a e^1.5 i x
but e^ix = cos x + i sin x
y = a e^kx
y' = a k e^kx
y" = a k^2 e^kx
4 a k^2 = 9 a
k = +/- 3/2
more interesting would be
4 y" + 9 y = 0
then you would get
k^2 = -9/4
and k = 1.5 i
so
y = a e^1.5 i x
but e^ix = cos x + i sin x
you know it will be some exponentials e^ax where the exponents solve 4a^2-9 = 0
y = c1 e^(3/2 x) + c2 e^(-3/2 x)
To find the complete general solution to the second-order differential equation 4y'' - 9y = 0, we can follow these steps:
Step 1: Write down the characteristic equation.
The characteristic equation is obtained by substituting y = e^(rt) into the differential equation, where r is an unknown constant. In this case, the characteristic equation is 4r^2 - 9 = 0.
Step 2: Solve the characteristic equation.
To solve the quadratic equation 4r^2 - 9 = 0, we can factor it as (2r + 3)(2r - 3) = 0. This implies that r = -3/2 or r = 3/2.
Step 3: Write down the general solution.
Since we have two distinct roots, we will have two linearly independent solutions. The general solution to the differential equation is of the form y(t) = c1 * e^(-3/2t) + c2 * e^(3/2t), where c1 and c2 are constants.
This is the complete general solution to the second-order differential equation 4y'' - 9y = 0.