Solve y'' - y = 2e^(2x), y(0) = 1, y'(0) = 0

Step by step

The question

Question: solve y″ − y = 2e^(2x), with y(0) = 1 and y′(0) = 0.

The answer has two parts: the plain part, which solves the equation with 0 on the right, plus one extra solution that makes 2e^(2x).

1 Solve with 0 on the right

Try y = e^(rx): the characteristic equation is r² − 1 = 0.

Two numbers that multiply to −1 and add to 0: 1 and −1.

So it factors: (r + 1)(r − 1) = 0.

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

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 C₁e^(−x) + C₂eˣ.

2 Guess the extra part

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

So we guess Ae^(2x) 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 = 2/3.

So the extra part is 2/3 × e^(2x).

5 Add the parts

Add the parts first.

The starting values are for the whole y.

6 Use the starting values

We need y′ too: the derivative of our y.

Put x = 0 into y: C₁ + C₂ + 666667/1000000 = 1.

Move the 666667/1000000 over.

Put x = 0 into y′: −C₁ + C₂ + 1333333/1000000 = 0.

Move the 1333333/1000000 over.

From ①: C₁ = 333333/1000000 − C₂.

Put it into ②.

Open the bracket and collect: −1 × 333333/1000000 moves over.

So C₂ = −1/2, and C₁ = 833333/1000000.

Put them in: y = 833333/1000000 × e^(−x) − ½eˣ + 2/3 × e^(2x).

7 Check

Let's check the extra part: put 2/3 × e^(2x) into the left side.

It gives 2e^(2x): the right side.

Right!

And the starting values: 1 and 0.

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.

Then: use the starting values.

Last: check.