Notes about a few exercises

These are not intended to be solutions. Just some thoughts.

Chapter 1

• (1.1.105) You can use the method of example 1.1.4. Alternatively, note that this equation is separable.

• (1.2.102) This equation is separable. The general solution of \(y' = xy\) is of the form \(y = Ce^{x^2/2}\) for some constant \(C\).

• (1.2.103) Solving this differential equation (ignoring the initial value condition) just involves taking an antiderivative. You can take this antiderivative using partial fractions.

• (1.2.105) Note that, if the initial condition was of the form \(y(0) = a\) for some \(a > 0\), then the constant function \(y \equiv a\) is a solution. If the initial condition was of the form \(y(0) = a\) for some \(a < 0\), then the function \(y = x + a\) is a solution in an interval containing \(x = 0\).

  Note also that, not only is there not a continuously differentiable solution when the initial condition is \(y(0) = 0\), there is just no solution whatsoever (more precisely, I mean that there is no differentiable function defined on an interval containing \(x = 0\) that satisfies the differential equation). This is a consequence of Darboux’s theorem. If you’ve taken some real analysis, I encourage you to work out the details and write up a formal \(\epsilon\)-\(\delta\) proof.

• (1.2.7) Possible hint: if \(g\) is a solution of the ODE \(y' = f(x,y)\), consider the function \(h\) defined by \(h(x) = g(x) - x\).

• (1.3.107) Possible hint: partial fractions.

• (1.4.104) This equation is linear, but it’s also separable. To get more practice, try solving it both ways!