Classify each ODE as linear or not linear:\(x’ + 3x = \cos(t)\): [resp_0]\(x(x’)=t\): [resp_1]\(\frac{dx}{dt}=t^3x\): [resp_2]\((x’)^3 + x = 0\): [resp_3]\(x’+\sqrt{t}\,x=0\): [resp_4]\(e^x x’ + x = t\): [resp_5]

A student finds the general solution to \(y’=-2y\) and gives \(y=Ce^{{-2t}}\). What is the issue?

Suppose for a population P(t): the birth rate per person is proportional to P with proportionality constant 4; the death rate per person is proportional to \(P^2\) with proportionality constant 2. Write the ODE for \(dP/dt\).

An autonomous ODE has equilibria at \(P=5\) (unstable) and \(P=10\) (stable). If \(P(0)=7\), what does P(t) tend to as \(t\to\infty\)?

The Embargo Act of 1806  by Jefferson was most criticized by the Federalists

Solve \(\frac{dv}{dt}=\frac{4t}{2v}\) where \(v(0)=-3\)

The acceleration of a particle equals the negative of its position x(t). Write the ODE for x(t).

Consider \(y’=2(y-1)^2\). After separation of variables we get the solutions \(y=1-\frac{{1}}{{2x+C}}\). Find the solution satisfying \(y(0)=1\).

An autonomous ODE has equilibria at \(P=5\) (unstable) and \(P=10\) (stable). If \(P(0)=10\), what is P(t)?

Goal: solve \(y’+\frac{{2}}{{x}}y=x\). Integrating factor \(\rho=x^2\) gives: $$(x^2y)’=x^3$$ After integrating we get \(x^2y=\frac{{x^4}}{{4}}+C\). What is the correct solution for y?