Solve y'' + y = 3e^(2x)

Step by step

The question

Question: solve y″ + y = 3e^(2x).

The answer has two parts: the plain part, which solves the equation with 0 on the right, plus one extra solution that makes 3e^(2x).

1 Solve with 0 on the right

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

It does not factor nicely, so we use the formula, with a = 1, b = 0, c = 1.

Put the numbers in.

Under the root: −4.

A minus under the root: √(−4) = 2i, because i = √(−1).

Divide both parts by 2: r = ±i.

Two complex roots.

Complex roots give cosines and sines.

The real part 0 goes in the e, the 1 goes inside cos and sin.

So the plain part is C₁cos x + C₂sin x.

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 = 3/5.

So the extra part is 3/5 × e^(2x).

5 Add the parts

Add the parts: y = C₁cos x + C₂sin x + 3/5 × e^(2x).

6 Check

Let's check the extra part: put 3/5 × e^(2x) into the left side.

It gives 3e^(2x): 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.