Solve y'' - 5y' + 6y = 12 Step by Step | Undetermined coefficients

Step by step

The question

Question: solve y″ − 5y′ + 6y = 12.

y″ − 5y′ + 6y = 12

The answer has two parts: the plain part, which solves the equation with 0 on the right, plus one extra solution that makes 12.

y = y_c + y_p: the plain part + one extra solution

1 Solve with 0 on the right

Try y = e^(rx): the characteristic equation is r² − 5r + 6 = 0.

r² − 5r + 6 = 0

y″ → r², y′ → r, y → 1

Two numbers that multiply to 6 and add to −5: −2 and −3.

−2 × (−3) = 6, −2 + (−3) = −5

So it factors: (r − 2)(r − 3) = 0.

(r − 2)(r − 3) = 0

Each bracket can be 0: r = 2 or r = 3.

r = 2 or r = 3

Each root gives a solution e^(rx).

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

two roots r₁, r₂: C₁e^(r₁x) + C₂e^(r₂x)

So the plain part is C₁e^(2x) + C₂e^(3x).

y_c = C₁e^(2x) + C₂e^(3x)

2 Guess the extra part

The extra part has the same shape as the right side: just a number.

a number → A

So we guess A for the extra part.

A is the numbers we must find.

y_p = A

3 Put the guess in

Its derivative.

y′_p = 0

And once more.

y″_p = 0

Put them into the left side.

0 − 5(0) + 6 × A = 12

Multiply out and put like parts together.

6A = 12

4 Match the parts

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

number parts: 6A = 12

same shape → same number

So A = 2.

A = 2

So the extra part is 2.

y_p = 2

5 Add the parts

Add the parts: y = C₁e^(2x) + C₂e^(3x) + 2.

y = C₁e^(2x) + C₂e^(3x) + 2

6 Check

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

put in y_p: 0 − 5(0) + 6 × 2

It gives 12: the right side.

Right!

= 12

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.

Answer