Mathematics

13/3848–12FE Reference Handbook 10.6 · Mathematics · p. 53 First-order linear homogeneous DE

Handbook formula

First-Order Linear DE

\frac{d y}{d x} + P (x) y = Q (x), μ = e^{\int P d x}, y = \frac{1}{μ} (\int μ Q d x + C)

Handbook 10.6 p. 53 prints y′ + a y = 0 with solution y = C e^{−at}. It does not print a general integrating factor. μ = e^{∫P dx} below is the standard extension for y′ + P(x)y = Q(x), not a boxed handbook formula. The worked answer still follows that extension.

$P (x)$: Coefficient of y in standard form
$Q (x)$: Forcing function
$μ$: Integrating factor
$C$: Constant of integration

Step-by-step solved example

Solve dy/dx + 2y = 4 with y(0) = 1.

First-order linear DE: solution shape like exponential growth/decay.

1. Integrating factor
P = 2 (constant), so μ = e^{2x}.
$μ = e^{2 x}$
2. Exact derivative
d/dx (y e^{2x}) = 4 e^{2x} → y e^{2x} = 2 e^{2x} + C.
3. IC
y = 2 + C e^{-2x}. y(0)=1 → C = −1. So y = 2 − e^{-2x}.
$y = 2 - e^{- 2 x}$

Answer: y = 2 − e^{−2x}

10 practice questions

0/10 correct

1.The integrating factor for y' + 3y = x is

2.y' = ky (k constant) has solutions

3.For y' + y = 0, μ equals

4.The general solution of y' = 2x is

5.A first-order linear DE is already in standard form when it looks like

6.If μ = e^{∫P dx}, then d(μy)/dx equals

7.y' + (1/x) y = 0 (x>0). μ =

8.Steady state of y' + 2y = 4 as x→∞ is

9.Separable means you can write

10.y(0)=0 for y' − y = 0 implies