Solutions of Some Differential Equations

Jim LambersMAT 285Spring Semester 2012-13Lecture 3 NotesThese notes correspond to Section 1.2 in the text.Solutions of Some Differential EquationsBoth differential equations from the preceding examples are of the formdydt= ay − b,where a and b are constants. In this section, we seek a solution of this more general differentialequation. That is, we will find a function y(t) that satisfies the equation.We will see that a differential equation such as this one does not have a unique solution, as itdoes not include enough information. Typically, the differential equation is paired with an initialcondition of the formy(t 0 ) = y 0 ,where t 0 represents an initial time and y 0 is an initial value. The differential equation, togetherwith the intial condition, is called an initial value problem. Under certain assumptions, it can beproven that an initial value problem has a unique solution.We now solve the above differential equation. Rearranging yields1 dyy − (b/a) dt = a.Then, we can integrate both sides with respect to t:∫∫1 dyy − (b/a) dt dt =a dt.On the left side, we use a substitutionto simplify the integral, which yields∫y = y(t),dy = dydt dt∫1y − (b/a) dy =a dt.Integrating both sides, we obtainln∣ y − b a∣ = at + C,where C is an arbitrary constant. Exponentiating both sides yields∣ y − b a∣ = eat+C ,1

which can be rearranged to obtainy = b a + ceat ,where c = ±e C .This solution is known as the general solution. It is a function of the independent variable tas well as the parameter c, the value of which depends on the initial condition. Thus, the generalsolution consists of infinitely many curves, each one being the graph of a function of t correspondingto a different value of c. Each such curve is called an integral curve.We now use the general solution to obtain the solution of the initial value problem. Substitutingthe initial condition into the general solution, we haveWe then solve this equation for c to obtainc =y 0 = b a + ceat 0.(y 0 − b )e −at 0.aSubstituting this expression for c into the general solution yields the solution of the inital valueproblem,y = b (a + y 0 − b )e a(t−t0) .aExample We revisit the equationdvdt = 9.8 − v 5 .This is an equation of the form dy/dt = ay − b, with y = v, a = −1/5 and b = −9.8. Applying theabove formula for the general solution yieldsv(t) = 49 + ce −t/5 ,where c is an arbitrary constant. Now, if we add the initial condition v(0) = 50, we obtain theunique solutionv(t) = 49 + (50 − 49)e −(t−0)/5 = 49 + e −t/5 .✷2