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Differential Equations!

Differential Equations!

Chara Chara
January 8, 2026
3 min read
Math UCSBB
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Today’s note (Jan 8): first-order linear ODEs, integrating factors, and integration by parts, with every small step shown.

First-order linear ODE

dydx+P(x) y=Q(x),\frac{dy}{dx} + P(x)\, y = Q(x),

where P(x)P(x) and Q(x)Q(x) are known functions.

To solve it, multiply both sides by a function μ(x)\mu(x) so that the left-hand side becomes the derivative of a product.

Definition (Integrating factor)

For y′+P(x) y=Q(x)y' + P(x)\, y = Q(x), choose

μ(x)=e∫P(x) dx.\mu(x) = e^{\int P(x)\,dx}.

Using the chain rule (ef(x))′=ef(x)f′(x)(e^{f(x)})' = e^{f(x)} f'(x) and the fact that (∫P(x) dx)′=P(x)\bigl(\int P(x)\,dx\bigr)' = P(x), we get

(μ(x) y(x))′=μ(x)P(x)y(x)+μ(x)y′(x).(\mu(x)\, y(x))' = \mu(x) P(x) y(x) + \mu(x) y'(x).

Multiplying the original equation by μ(x)\mu(x) makes the left side collapse to a single derivative:

(μ(x) y(x))′=μ(x) Q(x).(\mu(x)\, y(x))' = \mu(x)\, Q(x).

Integrating both sides then gives

y(x)=1μ(x)(∫μ(x) Q(x) dx+C).y(x) = \frac{1}{\mu(x)}\left(\int \mu(x)\, Q(x)\, dx + C\right).

We deliberately pick μ(x)\mu(x) so that

μ(x)y′(x)+μ(x)P(x)y(x)=(μ(x)y(x))′,\mu(x) y'(x) + \mu(x) P(x) y(x) = (\mu(x) y(x))',

turning the left-hand side into one derivative.

Recall (Integration by parts refresher)

Use this when integrating products (like polynomials times exponentials):

∫u dv=u v−∫v du.\int u\, dv = u\, v - \int v\, du.

Equivalently, the product rule says ∫f′(x)g(x) dx+∫f(x)g′(x) dx=f(x)g(x)+C\int f'(x) g(x)\, dx + \int f(x) g'(x)\, dx = f(x) g(x) + C.

Example 1 — y′+2y=x e−xy' + 2y = x\, e^{-x}

Here P(x)=2P(x) = 2, so μ(x)=e∫2 dx=e2x\displaystyle \mu(x) = e^{\int 2\, dx} = e^{2x}.

Multiplying by e2xe^{2x}:

e2xy′+2e2xy=x ex.e^{2x} y' + 2 e^{2x} y = x\, e^{x}.

Since (e2xy)′=(e2x)′y+e2xy′=2e2xy+e2xy′\bigl(e^{2x} y\bigr)' = (e^{2x})' y + e^{2x} y' = 2 e^{2x} y + e^{2x} y', the left side is

(e2xy)′=x ex.\bigl(e^{2x} y\bigr)' = x\, e^{x}.

Integrate both sides:

e2xy(x)=∫x exdx.e^{2x} y(x) = \int x\, e^{x} dx.

Use integration by parts with u=xu = x, dv=exdxdv = e^{x} dx:

∫x exdx=xex−∫exdx=xex−ex+C=(x−1)ex+C.\int x\, e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x} + C = (x - 1) e^{x} + C.

Therefore

e2xy(x)=(x−1)ex+C,e^{2x} y(x) = (x - 1) e^{x} + C,

so

y(x)=(x−1)e−x+C e−2x.y(x) = (x - 1) e^{-x} + C\, e^{-2x}.

Example 2 — y′+1xy=x3(x>0)y' + \dfrac{1}{x} y = x^{3} \quad (x > 0)

Here P(x)=1xP(x) = \dfrac{1}{x}, so μ(x)=e∫1xdx=x\displaystyle \mu(x) = e^{\int \frac{1}{x} dx} = x.

Multiplying by xx gives

x y′+y=x4.x\, y' + y = x^{4}.

Check the product derivative: (xy)′=xy′+y(x y)' = x y' + y, so

(x y)′=x4.(x\, y)' = x^{4}.

Integrate:

x y=∫x4dx=x55+C,x\, y = \int x^{4} dx = \frac{x^{5}}{5} + C,

so

y(x)=x45+Cx.y(x) = \frac{x^{4}}{5} + \frac{C}{x}.

Example 3 — y′−y=e2xy' - y = e^{2x} with y(0)=1y(0) = 1

Here P(x)=−1P(x) = -1, so μ(x)=e∫−1 dx=e−x\displaystyle \mu(x) = e^{\int -1\, dx} = e^{-x}.

Multiply by e−xe^{-x}:

e−xy′−e−xy=ex.e^{-x} y' - e^{-x} y = e^{x}.

Using (e−xy)′=(e−x)′y+e−xy′=−e−xy+e−xy′(e^{-x} y)' = (e^{-x})' y + e^{-x} y' = -e^{-x} y + e^{-x} y', we get

(e−xy)′=ex.(e^{-x} y)' = e^{x}.

Integrate:

e−xy(x)=∫exdx=ex+C.e^{-x} y(x) = \int e^{x} dx = e^{x} + C.

So

y(x)=e2x+C ex.y(x) = e^{2x} + C\, e^{x}.

Apply the initial condition y(0)=1y(0) = 1:

1=e0+C e0⇒C=0,1 = e^{0} + C\, e^{0} \quad \Rightarrow \quad C = 0,

giving the particular solution

y(x)=e2x.y(x) = e^{2x}.