Solve y' + 2y = 4x + 3 with y(0) = 1 by an integrating factor, step by step

1

See what kind of equation this is, and what would make it easy

Why this step

The unknown here is a whole function $y(x)$ , and the equation ties its slope $y'$ to its own value. We cannot simply integrate both sides, because $y$ is tangled up on the left with its own derivative. The idea that unlocks it: the left side ALMOST looks like the result of the product rule. If it were exactly the derivative of a product, one integration would finish the job. So we ask what we could multiply by to make it so.

y' + 2y = 4 x + 3, \qquad y(0) = 1

y'

means the derivative of

y

with respect to

x

: the slope of the graph of

y

. The second statement is the starting value, which will single out one solution from a whole family.

The little strokes show the slope the equation demands at each point, which is 4x + 3 minus 2 times y. A solution is a curve that follows the strokes everywhere. The bright curve is the one that passes through the starting point.

2

Find the integrating factor

Why this step

Multiply the equation by an unknown function $\mu(x)$ (the Greek letter mu). The left side becomes $\mu y' + 2 \mu y$ . The product rule says $(\mu y)' = \mu y' + \mu' y$ . These two match exactly when $\mu' = 2 \mu$ : a function whose derivative is $2$ times itself. Exponentials are the functions that do that.

\mu(x) = e^{2 x}

The derivative of

e^{kx}

is

k e^{kx}

, so this

\mu

satisfies

\mu' = 2 \mu

. In general the integrating factor is

e^{\int p(x) \, dx}

, and here

\int 2 \, dx = 2 x

.

The little strokes show the slope the equation demands at each point, which is 4x + 3 minus 2 times y. A solution is a curve that follows the strokes everywhere. The bright curve is the one that passes through the starting point.

3

Multiply through and recognise a product rule

Why this step

This is the moment the method was designed for. After multiplying by $\mu$ , the left side is no longer two tangled terms; it is the derivative of the single product $\mu y$ , and a derivative is something we know how to undo.

e^{2 x} y' + 2 e^{2 x} y = e^{2 x} \left( 4 x + 3 \right)

Every term multiplied by

e^{2 x}

.

\left( e^{2 x} \, y \right)' = e^{2 x} \left( 4 x + 3 \right)

The left side is exactly what the product rule gives for

\left( e^{2 x} y \right)'

: first times derivative of second, plus derivative of first,

2 e^{2 x}

, times second.

The little strokes show the slope the equation demands at each point, which is 4x + 3 minus 2 times y. A solution is a curve that follows the strokes everywhere. The bright curve is the one that passes through the starting point.

4

Integrate both sides

Why this step

Integrating the left side simply removes the derivative and leaves $\mu y$ . All the real work is on the right: we need a function whose derivative is $e^{2 x} \left( 4 x + 3 \right)$ . A polynomial times an exponential calls for integration by parts, which is the product rule run backwards: it trades the integral for an easier one in which the polynomial has been differentiated away.

Take $u = 4 x + 3$ , so $du = 4 \, dx$ ; and $dv = e^{2 x} \, dx$ , so $v = \frac{1}{2} e^{2 x}$ . Then:

\int e^{2 x} \left( 4 x + 3 \right) dx = \frac{1}{2} e^{2 x} \left( 4 x + 3 \right) - \int \left( 2 e^{2 x} \right) \, dx

u v

minus the integral of

v \, du

. The remaining integral has only an exponential in it.

= \frac{1}{2} e^{2 x} \left( 4 x + 3 \right) - e^{2 x} + C

\int e^{kx} dx = \frac{1}{k} e^{kx}

, and now the constant of integration

C

appears: it is what will later absorb the starting value.

2 e^{2 x} \left( 2 x + \frac{1}{2} \right) + 2 e^{2 x} = e^{2 x} \left( 4 x + 3 \right)

Check the integral by differentiating the answer,

e^{2 x} \left( 2 x + \frac{1}{2} \right)

, with the product rule. It returns the function we integrated. (Evaluated at several values of

x

.) checked

e^{2 x} \, y = e^{2 x} \left( 2 x + \frac{1}{2} \right) + C

Both sides integrated, with the right side collected into one bracket.

The little strokes show the slope the equation demands at each point, which is 4x + 3 minus 2 times y. A solution is a curve that follows the strokes everywhere. The bright curve is the one that passes through the starting point.

5

Divide by the integrating factor

Why this step

We introduced $\mu$ only as scaffolding. Dividing by it, which is safe because an exponential is never zero, leaves $y$ on its own. What appears is a whole family of solutions, one for each value of $C$ , and the two parts of it mean different things: one part is forced by the right side of the equation, the other is the free response of the equation itself.

y = 2 x + \frac{1}{2} + C e^{-2 x}

Dividing

C

by

e^{2 x}

gives

C e^{-2 x}

. This is the general solution.

The little strokes show the slope the equation demands at each point, which is 4x + 3 minus 2 times y. A solution is a curve that follows the strokes everywhere. The bright curve is the one that passes through the starting point.

6

Use the starting value to find the constant

Why this step

A first order equation has one free constant, and one piece of information fixes it. Without the starting value the equation describes every possible history of the system; the starting value picks out the one that actually happened.

\displaystyle y(0)

=

\displaystyle \frac{1}{2} + C

Put

x = 0

into the general solution.

e^0 = 1

, so the exponential disappears. checked

=

\displaystyle 1

This must equal the starting value, which fixes

C = 1 - \frac{1}{2} = \frac{1}{2}

. checked

y = 2 x + \frac{1}{2} + \frac{1}{2} e^{-2 x}

The solution of the starting value problem.

The little strokes show the slope the equation demands at each point, which is 4x + 3 minus 2 times y. A solution is a curve that follows the strokes everywhere. The bright curve is the one that passes through the starting point.

7

Check the answer in the original equation

Why this step

A differential equation hands us the perfect test of any proposed answer: differentiate it, put it in, and see whether the two sides agree for EVERY $x$ . No trust in the method is required; the check would expose a slip in any of the steps above.

y' = 2 - e^{-2 x}

Differentiate the solution. The derivative of

e^{k(x - x_0)}

is

k e^{k(x - x_0)}

.

2 - e^{-2 x} + 2 \left( 2 x + \frac{1}{2} + \frac{1}{2} e^{-2 x} \right) = 4 x + 3

y' + 2 y

with our

y

put in. The exponentials cancel and the right side of the equation appears. (Evaluated at several values of

x

.) checked

y(0) = 1

The starting value. checked

The little strokes show the slope the equation demands at each point, which is 4x + 3 minus 2 times y. A solution is a curve that follows the strokes everywhere. The bright curve is the one that passes through the starting point.

First order linear equation with an integrating factor