Question

Find the second solution of differential equation y'' + y = 0

Answer 1

y' + y = 0 is a second order first degree differential equation in two variables x and y.

Let the solution be in the form of
y = A Sin (w x + θ)

where A , w and θ are constants.

So y' = A w Cos (w x + θ) by differentiating the y.

Again differentiating we get
y'' = - A w^2 Sin (w x + θ) = - A w^2 y

Comparing with the given differential equation we get A w^2 = 1.

A = 1/w^2.

The solution is y = 1/w^2 × Sin (w x + θ)

The constants w and θcyan be determined by initial conditions if they are known.

Answer 2

Answer:

Step-by-step explanation:

Let the solution be in the form of

y = A Sin (w x + θ)

where A , w and θ are constants.

So y' = A w Cos (w x + θ) by differentiating the y.

Again differentiating we get

y'' = - A w^2 Sin (w x + θ) = - A w^2 y

Comparing with the given differential equation we get A w^2 = 1.

A = 1/w^2.

The solution is y = 1/w^2 × Sin (w x + θ)

The constants w and θcyan be determined by initial conditions if they are known