(New page: Back to the MA366 course wiki == Separable Equations == If the differential equation in question can be written as <math>M(t)dt=N(y)dy</math> then the equation is called sepa...) |
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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. | 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. | ||
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| + | 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. | ||
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| + | Suppose we are given the differential equation: | ||
| + | |||
| + | <math>yy'+\frac{1}{t^3}=\frac{5}{t}</math> | ||
| + | |||
| + | This is one of those times where you should use the notation dy/dt instead of just y'. So let's use that notation. | ||
| + | |||
| + | <math>y\frac{dy}{dt}+\frac{1}{t^3}=\frac{5}{t}</math> | ||
| + | |||
| + | 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: | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | y\frac{dy}{dt}+\frac{1}{t^3}&=\frac{5}{t}\\ | ||
| + | y\frac{dy}{dt}&=\frac{5}{t}-\frac{1}{t^3} | ||
| + | \end{align} | ||
| + | </math> | ||
Revision as of 10:09, 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} $
This is one of those times where you should use the notation dy/dt 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} \end{align} $
