Is y = e^(4x) a solution of y'' - 5y' + 6y = 0 Step by Step

Step by step

The question

Question: check if y = e^(4x) solves y″ − 5y′ + 6y = 0.

y = e^(4x)

y″ − 5y′ + 6y = 0 is a differential equation: an equation with a derivative in it.

y″ − 5y′ + 6y = 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^(4x) is 4e^(4x).

y′ = 4e^(4x)

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

(e^(kx))′ = ke^(kx)

Careful!

The chain rule!

Times the inside's slope, 4.

Careful: not (e^(4x))′ = e^(4x) ✗

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

It is 16e^(4x).

y″ = 16e^(4x)

2 Put them in

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

16e^(4x) − 5 × 4e^(4x) + 6 × e^(4x) = 0

Multiply out and add like terms.

2e^(4x) = 0

3 Compare the sides

The sides are different.

One number shows it: at x = 1, the left side is 109.2 and the right side is 0.

x = 1: left = 109.2, right = 0, 109.2 ≠ 0

So y = e^(4x) is not a solution.

= no: y = e^(4x) is not a solution

4 Check

Let's check with another number, x = 0.5: 14.78 and 0.

Different again.

Right!

x = 0.5: left ≈ 118.22 − 147.78 + 44.33 = 14.78, right ≈ 0

Let's look back.

First: find the derivatives.

Then: put them in.

Then: compare the sides.

Last: check.

Answer