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Integrate on both sides <math>\int\frac{1}{y}dy=\int-2tdt</math> and get <math>ln|y|=-t^2+C</math>, where <math>C</math> is a constant. | Integrate on both sides <math>\int\frac{1}{y}dy=\int-2tdt</math> and get <math>ln|y|=-t^2+C</math>, where <math>C</math> is a constant. | ||
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| + | Reconstruct this equation, <math>y=e^-t^2+C</math> | ||
</font> | </font> | ||
Revision as of 21:11, 12 November 2017
Basic Methods to Solve 1st-Order ODEs
3.0 Abstract
By now we have known what is a differential equation and how its solutions conduct. It's time to solve it, like plenty of linear equations we have done before.
3.1 Separable Equation
The easiest method is to separate the variables. This method is switching the variables to make the same variable on the same side, in order to integral on both sides and solve out the function (solution).
For example, we want to solve the differential equation $ \frac{dy}{dt}=-2yt $, where $ y(0)=1 $.
Now we start separating our variables. We put all items with respect to our dependent variable $ y $ on the left hand side of the equation, and all with respect to our independent variable $ t $ on the right hand side. Hence we get $ \frac{1}{y}dy=-2tdt $.
Integrate on both sides $ \int\frac{1}{y}dy=\int-2tdt $ and get $ ln|y|=-t^2+C $, where $ C $ is a constant.
Reconstruct this equation, $ y=e^-t^2+C $
