Skip to content

Mathematics · Ch 9 — Differential Equations

Linear Differential Equations

9.4.3

Linear Differential Equations

9.4.3 Linear Differential Equations

What Makes a Differential Equation Linear?

A differential equation is linear when the dependent variable and all its derivatives appear only in the first power and are not multiplied together. The standard form of a first order linear differential equation is:

dydx+Py=Q\frac{dy}{dx} + P y = Q

where PP and QQ are constants or functions of xx only. The key point is that yy appears alone — never as y2y^2, sin⁡y\sin y, or ydydxy\frac{dy}{dx}.

Note

The coefficient of dydx\frac{dy}{dx} must be 1. If it isn't, divide the entire equation by that coefficient first.

EquationPPQQ
dydx+y=sin⁡x\frac{dy}{dx} + y = \sin x11sin⁡x\sin x
dydx+yx=ex\frac{dy}{dx} + \frac{y}{x} = e^x1x\frac{1}{x}exe^x
dydx+yxlog⁡x=1x\frac{dy}{dx} + \frac{y}{x \log x} = \frac{1}{x}1xlog⁡x\frac{1}{x \log x}1x\frac{1}{x}

The Alternative Form: When xx is the Dependent Variable

Sometimes it is more convenient to treat xx as the dependent variable and yy as the independent variable:

dxdy+P1x=Q1\frac{dx}{dy} + P_1 x = Q_1

where P1P_1 and Q1Q_1 are constants or functions of yy only.

EquationP1P_1Q1Q_1
dxdy+x=cos⁡y\frac{dx}{dy} + x = \cos y11cos⁡y\cos y
dxdy−2xy=y2e−y\frac{dx}{dy} - \frac{2x}{y} = y^2 e^{-y}−2y-\frac{2}{y}y2e−yy^2 e^{-y}

The Integrating Factor

The method for dydx+Py=Q\frac{dy}{dx} + Py = Q multiplies both sides by a specially chosen function g(x)g(x) so that the left-hand side becomes the derivative of a product. Multiplying by g(x)g(x):

g(x)dydx+P⋅g(x)⋅y=Q⋅g(x)g(x) \frac{dy}{dx} + P \cdot g(x) \cdot y = Q \cdot g(x)

We want the left side to equal ddx[y⋅g(x)]=g(x)dydx+y⋅g′(x)\frac{d}{dx}[y \cdot g(x)] = g(x) \frac{dy}{dx} + y \cdot g'(x). Matching terms requires P⋅g(x)=g′(x)P \cdot g(x) = g'(x), i.e.

g′(x)g(x)=P  ⇒  log⁡∣g(x)∣=∫P dx  ⇒  g(x)=e∫P dx\frac{g'(x)}{g(x)} = P \;\Rightarrow\; \log |g(x)| = \int P \, dx \;\Rightarrow\; g(x) = e^{\int P \, dx}

Important

g(x)=e∫P dxg(x) = e^{\int P \, dx} is the Integrating Factor (I.F.) — the function that, when multiplied through, makes the left-hand side a perfect derivative.

The General Solution

Multiplying dydx+Py=Q\frac{dy}{dx} + Py = Q by the I.F., the left-hand side becomes ddx(ye∫P dx)\frac{d}{dx}\left( y e^{\int P \, dx} \right), so

ddx(ye∫P dx)=Qe∫P dx\frac{d}{dx}\left( y e^{\int P \, dx} \right) = Q e^{\int P \, dx}

Integrating gives the general solution ye∫P dx=∫Qe∫P dx dx+Cy e^{\int P \, dx} = \int Q e^{\int P \, dx} \, dx + C.

General Solution of dydx+Py=Q\frac{dy}{dx} + Py = Q

y⋅(I.F.)=∫Q⋅(I.F.) dx+C,I.F.=e∫P dxy \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) \, dx + C, \qquad \text{I.F.} = e^{\int P \, dx}

Procedure: write the equation in standard form and identify P,QP, Q; compute I.F.=e∫P dx\text{I.F.} = e^{\int P \, dx}; apply the formula above; then evaluate the integral and solve for yy.

Tip

When computing ∫P dx\int P \, dx you need not add a constant of integration — any constant cancels when forming the I.F.

The Case dxdy+P1x=Q1\frac{dx}{dy} + P_1 x = Q_1 …

Start your 14-day free trial to unlock the full solution →
← Previous · 9.4.2Q17 — Which of the following is a homogeneous differential equation? (A) (4x…Next · 9.4.3 →Example 14 — Find the general solution of the differential equation (dy)/(dx) - y =…