Solve y'' - 3y' + 2y = 6e^x

Step by step

The question

Question: solve y″ − 3y′ + 2y = 6eˣ.

The answer has two parts: the plain part, which solves the equation with 0 on the right, plus one extra solution that makes 6eˣ.

1 Solve with 0 on the right

Try y = e^(rx): the characteristic equation is r² − 3r + 2 = 0.

Two numbers that multiply to 2 and add to −3: −1 and −2.

So it factors: (r − 1)(r − 2) = 0.

Each bracket can be 0: r = 1 or r = 2.

Each root gives a solution e^(rx).

Any mix of the two works, so we add them with two numbers, C₁ and C₂.

So the plain part is C₁eˣ + C₂e^(2x).

2 Guess the extra part

The extra part has the same shape as the right side: a number times the same e power.

But eˣ is already in the plain part: it gives 0 on the left.

So we multiply it by x.

Careful!

That one is already in the plain part, so it can only give 0.

Times x.

So we guess Ax eˣ for the extra part.

A is the numbers we must find.

3 Put the guess in

Its derivative.

And once more.

Put them into the left side.

Multiply out and put like parts together.

4 Match the parts

The two sides must match, part by part: the same amount of each shape.

So A = −6.

So the extra part is −6x eˣ.

5 Add the parts

Add the parts: y = C₁eˣ + C₂e^(2x) − 6x eˣ.

6 Check

Let's check the extra part: put −6x eˣ into the left side.

It gives 6eˣ: the right side.

Right!

Let's look back.

First: solve with 0 on the right.

Then: guess the extra part.

Then: put the guess in.

Then: match the parts.

Then: add the parts.

Last: check.