| Line 15: | Line 15: | ||
<math>yy'+\frac{1}{t^3}=\frac{5}{t}</math> | <math>yy'+\frac{1}{t^3}=\frac{5}{t}</math> | ||
| − | + | If you think you've got a separable equation, then you'll need to use the dy/dt notation instead of just y'. So let's use that notation. | |
<math>y\frac{dy}{dt}+\frac{1}{t^3}=\frac{5}{t}</math> | <math>y\frac{dy}{dt}+\frac{1}{t^3}=\frac{5}{t}</math> | ||
| Line 24: | Line 24: | ||
\begin{align} | \begin{align} | ||
y\frac{dy}{dt}+\frac{1}{t^3}&=\frac{5}{t}\\ | y\frac{dy}{dt}+\frac{1}{t^3}&=\frac{5}{t}\\ | ||
| − | y\frac{dy}{dt}&=\frac{5}{t}-\frac{1}{t^3} | + | y\frac{dy}{dt}&=\frac{5}{t}-\frac{1}{t^3}\\ |
| + | y~dy&=(\frac{5}{t}-\frac{1}{t^3})dt | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
| + | |||
| + | We have reached our goal: the y's and dy's are on the left, and the t's and dt's are on the right. Once we have reached this state, we can integrate both sides of the equation: the left with respect to y, and the right with respect to t. | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | \int y~dy&=\int (\frac{5}{t}-\frac{1}{t^3})dt\\ | ||
| + | \frac{y^2}{2}&=5ln|t|+\frac{1}{2}t^{-2}+C\qquad \qquad (*) | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | And that's about it. But you should know a couple of things: first of all, the above expression describes y as an implicit function of x. In this example it is possible to solve for y, but this '''will not always be the case'''. Sometimes you're just stuck with a nasty implicit function. Secondly, in our solution, we could solve for y by taking the square root of both sides. This in and of itself is fine, but if this were an initial value problem, then a little danger would be lurking beneath the surface. Here's why. | ||
| + | |||
| + | Suppose the initial value we're given is y(1)=2. Plug the initial value into (*): | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | \frac{2^2}{2}&=5ln|1|+\frac{1}{2}1^{-2}+C\\ | ||
| + | 2&=0+\frac{1}{2}+C\\ | ||
| + | C&=\frac{3}{2} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Now, we plug C=3/2 back into (*): | ||
| + | |||
| + | <math> | ||
| + | \frac{y^2}{2}=5ln|t|+\frac{1}{2}t^{-2}+\frac{3}{2} | ||
| + | </math> | ||
| + | |||
| + | ... and solve for y. | ||
| + | |||
| + | <math> | ||
| + | y=\pm\sqrt{10ln|t|+t^{-2}+3} | ||
| + | </math> | ||
| + | |||
| + | Technically, we have two functions here: one corresponding to the positive root, and another corresponding to the negative root. Are both correct? Actually, no. Only one of them is. If we plug in the initial value y(1)=2, then we get | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | 2&=\pm\sqrt{0+1+3}\\ | ||
| + | 2&=\pm\sqrt{4}\\ | ||
| + | 2&=\pm2 | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Therefore, only the positive root is a solution to the initial value problem. So be aware of that danger. | ||
| + | |||
| + | [[MA366|Back to the MA366 course wiki]] | ||
Latest revision as of 09:56, 5 October 2009
Separable Equations
If the differential equation in question can be written as
$ M(t)dt=N(y)dy $
then the equation is called separable. The name refers to your ability to separate the function y and the dependent variable t, with all of the t's and dt's on one side and all of the y's and dy's on the other. If, by using a little bit of algebra, you can get your equation into this form, then you can simply integrate both sides to find the solution. Integrate the left side with respect to t, and the right side with respect to y, and your differential equation will turn into something more familiar and useful. Unfortunately, very few equations are separable, and you'll soon find yourself pining for them as you're forced to use much more involved methods of solution.
In Section 1, we used this fact in solving the initial value problems. In the differential equations I used, it was intuitively obvious that "integrating both sides" was allowed, but this won't always be the case.
Suppose we are given the differential equation:
$ yy'+\frac{1}{t^3}=\frac{5}{t} $
If you think you've got a separable equation, then you'll need to use the dy/dt notation instead of just y'. So let's use that notation.
$ y\frac{dy}{dt}+\frac{1}{t^3}=\frac{5}{t} $
The objective here is to get all of the y's and dy's on one side of the equation, and all t's and dt's on the other. As you know, dy/dt isn't actually a fraction, but you can sort of pretend it is and treat dy and dt as separate entities, and move them around with algebra. Let's do some algebra:
$ \begin{align} y\frac{dy}{dt}+\frac{1}{t^3}&=\frac{5}{t}\\ y\frac{dy}{dt}&=\frac{5}{t}-\frac{1}{t^3}\\ y~dy&=(\frac{5}{t}-\frac{1}{t^3})dt \end{align} $
We have reached our goal: the y's and dy's are on the left, and the t's and dt's are on the right. Once we have reached this state, we can integrate both sides of the equation: the left with respect to y, and the right with respect to t.
$ \begin{align} \int y~dy&=\int (\frac{5}{t}-\frac{1}{t^3})dt\\ \frac{y^2}{2}&=5ln|t|+\frac{1}{2}t^{-2}+C\qquad \qquad (*) \end{align} $
And that's about it. But you should know a couple of things: first of all, the above expression describes y as an implicit function of x. In this example it is possible to solve for y, but this will not always be the case. Sometimes you're just stuck with a nasty implicit function. Secondly, in our solution, we could solve for y by taking the square root of both sides. This in and of itself is fine, but if this were an initial value problem, then a little danger would be lurking beneath the surface. Here's why.
Suppose the initial value we're given is y(1)=2. Plug the initial value into (*):
$ \begin{align} \frac{2^2}{2}&=5ln|1|+\frac{1}{2}1^{-2}+C\\ 2&=0+\frac{1}{2}+C\\ C&=\frac{3}{2} \end{align} $
Now, we plug C=3/2 back into (*):
$ \frac{y^2}{2}=5ln|t|+\frac{1}{2}t^{-2}+\frac{3}{2} $
... and solve for y.
$ y=\pm\sqrt{10ln|t|+t^{-2}+3} $
Technically, we have two functions here: one corresponding to the positive root, and another corresponding to the negative root. Are both correct? Actually, no. Only one of them is. If we plug in the initial value y(1)=2, then we get
$ \begin{align} 2&=\pm\sqrt{0+1+3}\\ 2&=\pm\sqrt{4}\\ 2&=\pm2 \end{align} $
Therefore, only the positive root is a solution to the initial value problem. So be aware of that danger.
