6.1: The Simplest Type of Differential Equation
The simplest ordinary differential equations can be integrated directly by finding antiderivatives. These simplest odes have the form
\[\frac{d^{n} x}{d t^{n}}=G(t), \nonumber \]
where the derivative of \(x=x(t)\) can be of any order, and the right-hand-side may depend only on the independent variable \(t\) . As an example, consider a mass falling under the influence of constant gravity, such as approximately found on the Earth’s surface. Newton’s law, \(F=m a\) , results in the equation
\[m \frac{d^{2} x}{d t^{2}}=-m g, \nonumber \]
where \(x\) is the height of the object above the ground, \(m\) is the mass of the object, and \(g=9.8\) meter \(/ \sec ^{2}\) is the constant gravitational acceleration. As Galileo suggested, the mass cancels from the equation, and
\[\frac{d^{2} x}{d t^{2}}=-g \nonumber \]
Here, the right-hand-side of the ode is a constant. The first integration, obtained by antidifferentiation, yields
\[\frac{d x}{d t}=A-g t \nonumber \]
with \(A\) the first constant of integration; and the second integration yields
\[x=B+A t-\frac{1}{2} g t^{2} \nonumber \]
with \(B\) the second constant of integration. The two constants of integration \(A\) and \(B\) can then be determined from the initial conditions. If we know that the initial height of the mass is \(x_{0}\) , and the initial velocity is \(v_{0}\) , then the initial conditions are
\[x(0)=x_{0}, \quad \frac{d x}{d t}(0)=v_{0} . \nonumber \]
Substitution of these initial conditions into the equations for \(d x / d t\) and \(x\) allows us to solve for \(A\) and \(B\) . The unique solution that satisfies both the ode and the initial conditions is given by
\[x(t)=x_{0}+v_{0} t-\frac{1}{2} g t^{2} . \nonumber \]
For example, suppose we drop a ball off the top of a 50 meter building. How long will it take the ball to hit the ground? This question requires solution of Equation \ref{6.1} for the time \(T\) it takes for \(x(T)=0\) , given \(x_{0}=50\) meter and \(v_{0}=0\) . Solving for \(T\) ,
\[\begin{aligned} T &=\sqrt{\frac{2 x_{0}}{g}} \\ &=\sqrt{\frac{2 \cdot 50}{9.8}} \mathrm{sec} \\ & \approx 3.2 \mathrm{sec} . \end{aligned} \nonumber \]