Differential Equations - First Order: Separable Equations

Differential Equations - First Order: Separable Equations - ii

Example 4 Solve the following IVP and find the interval of validity of the solution.

y^{'} = e^{- y} (2 x - 4) y (5) = 0

This differential equation is easy enough to separate, so let's do that and then integrate both sides.

\begin{aligned} e^{y} d y & = (2 x - 4) d x \\ \int e^{y} d y & = \int (2 x - 4) d x \\ e^{y} & = x^{2} - 4 x + c \end{aligned}

Applying the initial condition gives

1 = 25 - 20 + c c = - 4

This then gives an implicit solution of.

e^{y} = x^{2} - 4 x - 4

We can easily find the explicit solution to this differential equation by simply taking the natural log of both sides.

y (x) = \ln (x^{2} - 4 x - 4)

Finding the interval of validity is the last step that we need to take. Recall that we can't plug negative values or zero into a logarithm, so we need to solve the following inequality

x^{2} - 4 x - 4 > 0

The quadratic will be zero at the two points

x = 2 \pm 2 \sqrt{2}

. A graph of the quadratic (shown below) shows that there are in fact two intervals in which we will get positive values of the polynomial and hence can be possible intervals of validity.

So, possible intervals of validity are

\begin{matrix} \infty < x < 2 - 2 \sqrt{2} \\ 2 + 2 \sqrt{2} < x < \infty \end{matrix}

From the graph of the quadratic we can see that the second one contains

x

= 5, the value of the independent variable from the initial condition. Therefore, the interval of validity for this solution is.

2 + 2 \sqrt{2} < x < \infty

Here is a graph of the solution.

Example 5 Solve the following IVP and find the interval of validity for the solution.

\frac{d r}{d θ} = \frac{r^{2}}{θ} r (1) = 2

This is actually a fairly simple differential equation to solve. We’re doing this one mostly because of the interval of validity.
So, get things separated out and then integrate.

\begin{aligned} \frac{1}{r^{2}} d r & = \frac{1}{θ} d θ \\ \int \frac{1}{r^{2}} d r & = \int \frac{1}{θ} d θ \\ - \frac{1}{r} & = \ln | θ | + c \end{aligned}

Now, apply the initial condition to find

c

- \frac{1}{2} = \ln (1) + c c = - \frac{1}{2}

So, the implicit solution is then,

- \frac{1}{r} = \ln | θ | - \frac{1}{2}

Solving for

r

gets us our explicit solution.

r = \frac{1}{\frac{1}{2} - \ln | θ |}

Now, there are two problems for our solution here. First, we need to avoid

θ = 0

because of the natural log. Notice that because of the absolute value on the

θ

we don’t need to worry about

θ

being negative. We will also need to avoid division by zero. In other words, we need to avoid the following points.

\begin{aligned} \frac{1}{2} - \ln | θ | & = 0 \\ \ln | θ | & = \frac{1}{2} exponentiate both sides \\ | θ | = e^{\frac{1}{2}} \\ θ & = \pm \sqrt{e} \end{aligned}

So, these three points break the number line up into four portions, each of which could be an interval of validity.

\begin{matrix} - \infty < θ < - \sqrt{e} \\ - \sqrt{e} < θ < 0 \\ 0 < θ < \sqrt{e} \\ \sqrt{e} < θ < \infty \end{matrix}

The interval that will be the actual interval of validity is the one that contains

θ = 1

. Therefore, the interval of validity is

0 < θ < \sqrt{e}

.
Here is a graph of the solution.

Example 6 Solve the following IVP.

\frac{d y}{d t} = e^{y - t} \sec (y) (1 + t^{2}) y (0) = 0

This problem will require a little work to get it separated and in a form that we can integrate, so let's do that first.

\begin{aligned} \frac{d y}{d t} & = \frac{e^{y} e^{- t}}{\cos (y)} (1 + t^{2}) \\ e^{- y} \cos (y) d y & = e^{- t} (1 + t^{2}) d t \end{aligned}

Now, with a little integration by parts on both sides we can get an implicit solution.

\begin{aligned} \int e^{- y} \cos (y) d y & = \int e^{- t} (1 + t^{2}) d t \\ \frac{e^{- y}}{2} (\sin (y) - \cos (y)) & = - e^{- t} (t^{2} + 2 t + 3) + c \end{aligned}

Applying the initial condition gives.

\frac{1}{2} (- 1) = - (3) + c c = \frac{5}{2}

Therefore, the implicit solution is.

\frac{e^{- y}}{2} (\sin (y) - \cos (y)) = - e^{- t} (t^{2} + 2 t + 3) + \frac{5}{2}

It is not possible to find an explicit solution for this problem and so we will have to leave the solution in its implicit form. Finding intervals of validity from implicit solutions can often be very difficult so we will also not bother with that for this problem.

As this last example showed it is not always possible to find explicit solutions so be on the lookout for those cases.

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Calculus III - 3-Dimensional Space: Equations of Lines