Differential equation?

Question

y" - 2y = 0

how do you do with second derivative?

Answer 1

y" - 2y = 0

This is a homogeneous DE.
We can use the Eigenvalue approach.
For each differential, we substitute a letter D.
This works because we are looking for solutions in exponential functions.

D^2 y - 2y = 0

Drop the y's.

D^2 - 2 = 0

D^2 = 2

D = +/- sqrt(2).

Sometimes, you will have to use the Quadratic formula to solve this equation. If the Eigenvalues come out as complex numbers, you will get oscillatory functions with sine and cosine terms.

The first solution is

y1 = e^(sqrt(2) x)

The second is

y2 = e^(-sqrt(2) x)

And the general solution is a linear combination of these

y = a e^(sqrt(2) x) + b e^(-sqrt(2) x)

where a and b are arbitrary constants.

Answer 2

This is an homogeneous linear differential equation. The solution is given by the roots of the algebraic equation r^2 - 2 = 0, whose solutions are r = sqrt(2) and r = -sqrt(2). So the solutions of the differential equation are

y = c1^e^(sqrt(2) x) + c2 e^(-sqrt(2)x)), where c1 and c2 are constants