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Differential Equations

6.4 First-Order Linear Differential Equations

Focus on standard linear form, solve an initial-value problem, long-term behavior.

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Ideas in this lesson

01Standard linear formWorked steps · five questions02Solve an initial-value problemWorked steps · five questions03Long-term behaviorWorked steps · five questions
Idea 1 of 3

Standard linear form

Divide first to obtain y′+P(x)y=Q(x)y' + P(x)y = Q(x). The integrating factor is exp⁡(∫Pdx)\exp (\int P dx).

Worked example 1

Choose an integrating factor for y′+2y=xy' + 2y = x.
  1. P(x)=2P(x) = 2
  2. ∫P(x)dx=2x\int P(x) dx = 2x
  3. μ(x)=exp⁡(∫Pdx)=e2x\mu (x) = \exp (\int P dx) = e^{2x}

Answer: e2xe^{2x}

Worked example 2

Choose an integrating factor for y′+(3x)y=1(x>0)y' + (\dfrac{3}{x})y = 1 (x > 0).
  1. P(x)=3xP(x) = \dfrac{3}{x}
  2. ∫P(x)dx=3ln⁡x\int P(x) dx = 3 \ln x
  3. μ(x)=exp⁡(∫Pdx)=x3\mu (x) = \exp (\int P dx) = x^{3}

Answer: x3x^{3}

Idea 2 of 3

Solve an initial-value problem

Multiplying by the integrating factor makes the left side a product derivative. Integrate, then determine the constant.

Worked example 1

Solve y′+y=2,y(0)=1y' + y = 2, y(0) = 1. Find y(0)y(0).
  1. μ(x)=ex\mu (x) = e^{x}; (exy)′=2ex(e^{x}y)' = 2e^{x}
  2. y=2+y = 2 + Ce −x^{-x}; y(0)=1  ⟹  C=−1y(0) = 1 \implies C = -1
  3. y(0)=1y(0) = 1

Answer: 11

Worked example 2

Solve y′+y=3,y(0)=2y' + y = 3, y(0) = 2. Find y(1)y(1).
  1. μ(x)=ex\mu (x) = e^{x}; (exy)′=3ex(e^{x}y)' = 3e^{x}
  2. y=3+y = 3 + Ce −x^{-x}; y(0)=2  ⟹  C=−1y(0) = 2 \implies C = -1
  3. y(1)=3+−1e=3+−1e≈2.63y(1) = 3 + \dfrac{-1}{e} = 3 + \dfrac{-1}{e} \approx 2.63

Answer: 3+−1e≈2.633 + \dfrac{-1}{e} \approx 2.63

Idea 3 of 3

Long-term behavior

Inspect the explicit solution as xx grows. A decaying transient can leave a nonzero equilibrium.

Worked example 1

Solve y′+y=2,y(0)=1y' + y = 2, y(0) = 1. Find y(0)y(0).
  1. μ(x)=ex\mu (x) = e^{x}; (exy)′=2ex(e^{x}y)' = 2e^{x}
  2. y=2+y = 2 + Ce −x^{-x}; y(0)=1  ⟹  C=−1y(0) = 1 \implies C = -1
  3. y(0)=1y(0) = 1

Answer: 11

Worked example 2

Solve y′+y=3,y(0)=2y' + y = 3, y(0) = 2. Find y(1)y(1).
  1. μ(x)=ex\mu (x) = e^{x}; (exy)′=3ex(e^{x}y)' = 3e^{x}
  2. y=3+y = 3 + Ce −x^{-x}; y(0)=2  ⟹  C=−1y(0) = 2 \implies C = -1
  3. y(1)=3+−1e=3+−1e≈2.63y(1) = 3 + \dfrac{-1}{e} = 3 + \dfrac{-1}{e} \approx 2.63

Answer: 3+−1e≈2.633 + \dfrac{-1}{e} \approx 2.63

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §6.4,p6.4, p. 424424. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.