1.2: Integrals as solutions

A first order ODE is an equation of the form

\[\dfrac{dy}{dx}=f(x,y) \nonumber \]

or just

\[y'=f(x,y) \nonumber \]

In general, there is no simple formula or procedure one can follow to find solutions. In the next few sections we will look at special cases where solutions are not difficult to obtain. In this section, let us assume that \(f\) is a function of \(x\) alone, that is, the equation is

\[y'=f(x) \label{1.1.1} \]

We could just integrate (antidifferentiate) both sides with respect to \(x\).

\[\int y' (x) dx = \int f(x) dx + C \nonumber \]

that is

\[y(x)=\int f(x) dx + C \nonumber \]

This \(y(x)\) is actually the general solution. So to solve Equation \(\ref{1.1.1}\), we find some antiderivative of \(f(x)\) and then we add an arbitrary constant to get the general solution.

Example \(\PageIndex{1}\)

Solve the initial value problem \(y' = 3x+2, \;\; y(0)=2\).

Solution

Example \(\PageIndex{4}\)

Find the general solution of \(y' = y^2\).

Solution

First we note that \(y = 0\) is a solution. We can now assume that \(y \ne 0\). Write

\[\dfrac{dx}{dy} = \dfrac{1}{y^2} \nonumber \]

We integrate to get

\[x = \dfrac{-1}{y} + C \nonumber \]

We solve for \(y = \dfrac{1}{C-x}\). So the general solution is

\[y = \dfrac{1}{C-x} \,\, or\,\, y=0 \nonumber \]

Note the singularities of the solution. If for example \(C=1\), then the solution as we approach \(x=1\). See Figure \(\PageIndex{1}\). Generally, it is hard to tell from just looking at the equation itself how the solution is going to behave. The equation \(y' = y^2\) is very nice and defined everywhere, but the solution is only defined on some interval \((-\infty, C)\) or \((C, \infty)\). Usually when this happens we only consider one of these the solution. For example if we impose a condition \(y(0) = 1\), then the solution is \(y=\frac{1}{1-x}\), and we would consider this solution only for \(x\) on the interval \((-\infty,1)\). In the figure, it is the left side of the graph.

Figure \(\PageIndex{1}\): Plot of \(y=\frac{1}{1-x}\).

Classical problems leading to differential equations solvable by integration are problems dealing with velocity, acceleration and distance. You have surely seen these problems before in your calculus class.

Example \(\PageIndex{5}\)

Suppose a car drives at a speed \(e^{t/2}\) meters per second, where \(t\) is time in seconds. How far did the car get in 2 seconds (starting at \(t = 0\))? How far in 10 seconds?

Solution

Let \(x\) denote the distance the car traveled. The equation is

\[x' = e^{t/2} \nonumber \]

We can just integrate this equation to get that

\[x(t) = 2e^{t/2} + C \nonumber \]

We still need to figure out \(C\). We know that when \(t = 0\), then \(x = 0\). That is, \(x(0) = 0\). So

\[ 0 = x(0) = 2 e^{0/2} + C = 2 + C \nonumber \]

Thus \(C = -2\) and

\[x(t) = 2 e^{t/2} - 2 \nonumber \]

Now we just plug in to get where the car is at 2 and at 10 seconds. We obtain

\[x(2) = 2e^{2/2} - 2 \approx 3.44 \text{~meters},~~~ x(10) = 2e^{10/2} - 2 \approx 294\text{~meters} \nonumber \]

Example \(\PageIndex{6}\)

Suppose that the car accelerates at a rate of \(t^2 \frac{m}{s^2}\). At time \(t = 0\) the car is at the 1 meter mark and is traveling at 10 ^m/_s. Where is the car at time \(t = 10\).

Solution

Well this is actually a second order problem. If \(x\) is the distance traveled, then \(x'\) is the velocity, and \(x''\) is the acceleration. The equation with initial conditions is

\[x'' = t^2, \quad x(0) = 1, \quad x'(0) = 10 \nonumber \]

What if we say \(x' = v\). Then we have the problem

\[v' = t^2, \quad v(0) = 10 \nonumber \]

Once we solve for \(v\), we can integrate and find \(x\).