Is y = cos(3x) a solution of y'' + 4y = 0 Step by Step

Step by step

The question

Question: check if y = cos(3x) solves y″ + 4y = 0.

y = cos(3x)

y″ + 4y = 0 is a differential equation: an equation with a derivative in it.

y″ + 4y = 0

Its answer is not a number.

It is a function y, and it must make the equation true for every x.

differential equation: an equation with a derivative

its answer: a function y, true for every x

1 Find the derivatives

The derivative of cos(3x) is −3sin(3x).

y′ = −3sin(3x)

The rules we use: the chain rule brings the number in front down.

(cos(bx))′ = −b sin(bx)

Careful!

The chain rule!

Times the inside's slope, 3.

Careful: not (cos(3x))′ = −sin(3x) ✗

Once more: y″ is the derivative of y′.

It is −9cos(3x).

y″ = −9cos(3x)

2 Put them in

Now put y, y′ and y″ into the equation, in place of the letters.

−9cos(3x) + 4 × cos(3x) = 0

Multiply out and add like terms.

−5cos(3x) = 0

3 Compare the sides

The sides are different.

One number shows it: at x = 1, the left side is 4.95 and the right side is 0.

x = 1: left = 4.95, right = 0, 4.95 ≠ 0

So y = cos(3x) is not a solution.

= no: y = cos(3x) is not a solution

4 Check

Let's check with another number, x = 0.5: −0.35 and 0.

Different again.

Right!

x = 0.5: left ≈ −0.64 + 0.28 = −0.35, right ≈ 0

Let's look back.

First: find the derivatives.

Then: put them in.

Then: compare the sides.

Last: check.

Answer