Review / Part I — Explicit & Qualitative Methods / Ch 1

Initial Value Problems

Ch 1 — Part I — Explicit & Qualitative Methods

An initial value problem (IVP) pairs an ODE with enough data at one point to pin down a specific solution. For a first-order ODE you need the value \(y(t_0)\); for an \(n\)-th order equation you need \(y(t_0), y'(t_0), \ldots, y^{(n-1)}(t_0)\).

Why this many conditions?

A first-order ODE \(y' = f(t,y)\) specifies the slope at each \((t,y)\). One initial point picks out one curve from the slope field. An \(n\)-th order linear ODE has an \(n\)-dimensional space of solutions, so \(n\) independent conditions are needed.

Solving an IVP

Find the general solution (involving arbitrary constants), then impose the initial conditions to pin down each constant.

Warning: nonlinear IVPs can misbehave

Existence and uniqueness do not always hold for nonlinear problems — solutions can blow up in finite time or fail to be unique. See the existence/uniqueness page.

Practice Problems

Problem 1easy
Solve the IVP \(y' = 3y,\; y(0) = 4\).
Hint
General solution is \(y(t) = Ce^{3t}\).
Solution

Plug in \(t = 0\): \(y(0) = C = 4\). So \(y(t) = 4 e^{3t}\).

Answer: \(y(t) = 4 e^{3t}.\)

Problem 2easy
Given general solution \(y(t) = c_1 e^t + c_2 e^{-t}\) of \(y''-y=0\), solve \(y(0) = 2,\; y'(0) = 0\).
Hint
Compute \(y'\) and plug in.
Solution

\(y(0) = c_1 + c_2 = 2,\; y'(0) = c_1 - c_2 = 0\Rightarrow c_1 = c_2 = 1\).

Answer: \(y(t) = e^t + e^{-t} = 2\cosh(t).\)

Problem 3easy
For the exponential model \(y'(t) = ry(t)\) with \(y(t_0) = y_0\), write the closed-form solution.
Hint
Use the time-translation trick.
Solution

General solution \(y(t) = Ce^{rt}\); imposing \(y(t_0) = y_0\) gives \(C = y_0 e^{-rt_0}\), so \(y(t) = y_0 e^{r(t-t_0)}\).

Answer: \(y(t) = y_0 e^{r(t-t_0)}.\)

Problem 4easy
If \(y_1(t), y_2(t)\) both solve a linear second-order IVP with the *same* initial conditions, what theorem do we need to conclude \(y_1 \equiv y_2\)?
Hint
Uniqueness.
Solution

The existence and uniqueness theorem for linear ODEs with continuous coefficients guarantees equality.

Answer: Uniqueness theorem.