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

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The problem

Solve this starting value problem.

y' + y = -1, \qquad y(0) = -2

The result

y = -1 - e^{-x}

Found with the integrating factor $e^{x}$ and checked in the original equation.

checked 5 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.

Picture 1 of 2Step 1: See what kind of equation this is, and what would make it easy

The little strokes show the slope the equation demands at each point, which is -1 minus 1 times y. A solution is a curve that follows the strokes everywhere. The bright curve is the one that passes through the starting point.

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' + y = -1, \qquad y(0) = -2

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.

Find the integrating factor

Why this step

Multiply the equation by an unknown function $\mu(x)$ (the Greek letter mu). The left side becomes $\mu y' + \mu y$ . The product rule says $(\mu y)' = \mu y' + \mu' y$ . These two match exactly when $\mu' = 1 \mu$ : a function whose derivative is $1$ times itself. Exponentials are the functions that do that.

\mu(x) = e^{x}

The derivative of

e^{kx}

k e^{kx}

, so this

\mu

satisfies

\mu' = 1 \mu

. In general the integrating factor is

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

, and here

\int 1 \, dx = x

First order linear equation with an integrating factor

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

Find the integrating factor

Multiply through and recognise a product rule

Integrate both sides

Divide by the integrating factor

Use the starting value to find the constant

Check the answer in the original equation