Review / Part II — Higher-Order & Nonlinear / Ch 10

Inhomogeneous First-Order Linear ODEs

Ch 10 — Part II — Higher-Order & Nonlinear

A first-order linear ODE \(y' + p(t) y = g(t)\) is solved by the integrating factor \(\mu(t) = e^{\int p(t) dt}\).

Recipe

  1. Compute \(\mu(t) = e^{\int p(t) dt}\) (no \(+C\) needed).
  2. Multiply: \((\mu y)' = \mu g\).
  3. Integrate: \(\mu y = \int \mu g \, dt + C\).
  4. Solve: \(y(t) = \mu^{-1}\Big(\int \mu g\, dt + C\Big)\).

Why it works

By the product rule, \((\mu y)' = \mu' y + \mu y' = \mu p y + \mu y' = \mu(y' + py)\).

Structure: homogeneous + particular

The solution splits as \(y_h + y_p\) where \(y_h = C/\mu\) (homogeneous kernel) and \(y_p\) is any particular solution.

Practice Problems

Problem 1medium
Solve \(y' + 2y = e^{-t},\; y(0) = 1\).
Hint
\(\mu = e^{2t}\).
Solution

\((e^{2t} y)' = e^t\Rightarrow e^{2t} y = e^t + C\Rightarrow y = e^{-t} + Ce^{-2t}\). From \(y(0)=1\): \(1 = 1 + C\Rightarrow C=0\).

Answer: \(y(t) = e^{-t}.\)

Problem 2medium
Solve \(ty' + 2y = t^3\) for \(t>0\).
Hint
Divide by \(t\): \(y' + (2/t) y = t^2\). Then \(\mu = e^{\int 2/t \, dt} = t^2\).
Solution

\((t^2 y)' = t^4\Rightarrow t^2 y = t^5/5 + C\Rightarrow y = t^3/5 + C/t^2\).

Answer: \(y(t) = t^3/5 + C/t^2.\)

Problem 3easy
Solve \(y' - y = t\).
Hint
\(\mu = e^{-t}\).
Solution

\((e^{-t} y)' = t e^{-t}\). Integrating by parts: \(\int t e^{-t} dt = -t e^{-t} - e^{-t} + C\). So \(e^{-t} y = -(t+1)e^{-t} + C\Rightarrow y = -(t+1) + C e^t\).

Answer: \(y(t) = C e^t - t - 1.\)