Is y = e^(-x)sin(2x) a solution of y'' + 2y' + 5y = 0 Step by Step

Step by step

The question

Question: check if y = e^(−x)sin(2x) solves y″ + 2y′ + 5y = 0.

y = e^(−x)sin(2x)

y″ + 2y′ + 5y = 0 is a differential equation: an equation with a derivative in it.

y″ + 2y′ + 5y = 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 e^(−x)sin(2x) is −e^(−x)sin(2x) + 2e^(−x)cos(2x).

y′ = −e^(−x)sin(2x) + 2e^(−x)cos(2x)

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

(e^(kx))′ = ke^(kx)

(sin(bx))′ = b cos(bx)

Careful!

The chain rule!

Times the inside's slope, −1.

Careful: not (e^(−x))′ = e^(−x) ✗

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

It is −3e^(−x)sin(2x) − 4e^(−x)cos(2x).

y″ = −3e^(−x)sin(2x) − 4e^(−x)cos(2x)

2 Put them in

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

−3e^(−x)sin(2x) − 4e^(−x)cos(2x) + 2(−e^(−x)sin(2x) + 2e^(−x)cos(2x)) + 5 × e^(−x)sin(2x) = 0

Multiply out and add like terms.

On the left, everything cancels: 0.

0 = 0

3 Compare the sides

Both sides are the same, for every x.

So y = e^(−x)sin(2x) is a solution.

= yes: y = e^(−x)sin(2x) is a solution

4 Check

Let's check with one number, x = 1.

A calculator gives 0 on both sides.

Right!

x = 1: left ≈ −0.39 − 1.28 + 1.67 = 0, right ≈ 0

Let's look back.

First: find the derivatives.

Then: put them in.

Then: compare the sides.

Last: check.

Answer