Solve y'' + 4y = 8x, y(0) = 1, y'(0) = 4

Step by step

The question

Question: solve y″ + 4y = 8x, with y(0) = 1 and y′(0) = 4.

The answer has two parts: the plain part, which solves the equation with 0 on the right, plus one extra solution that makes 8x.

1 Solve with 0 on the right

Try y = e^(rx): the characteristic equation is r² + 4 = 0.

It does not factor nicely, so we use the formula, with a = 1, b = 0, c = 4.

Put the numbers in.

Under the root: −16.

A minus under the root: √(−16) = 4i, because i = √(−1).

Divide both parts by 2: r = ±2i.

Two complex roots.

Complex roots give cosines and sines.

The real part 0 goes in the e, the 2 goes inside cos and sin.

So the plain part is C₁cos(2x) + C₂sin(2x).

2 Guess the extra part

The extra part has the same shape as the right side: a polynomial of the same size, with every lower power too.

Careful!

Every lower power is needed too: the derivatives make them.

So we guess Ax + B for the extra part.

A and B are 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.

So B = 0.

So the extra part is 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 = 1.

So C₁ is known at once.

Put x = 0 into y′: 2C₂ + 2 = 4.

Move the 2 over.

So C₂ = 1.

Put them in: y = cos(2x) + sin(2x) + 2x.

7 Check

Let's check the extra part: put 2x into the left side.

It gives 8x: the right side.

Right!

And the starting values: 1 and 4.

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.