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Differential equation (view source)
Revision as of 18:26, February 28, 2009
, 18:26, February 28, 2009no edit summary
<math> \frac{dy}{dx} + F(x)y = G(x) </math>
<math> \frac{dy}{dx} + F(x)y = G(x) </math>
To solve this type of differential equation, an ''integrating factor''<ref>Edwards, Henry C. and Penney, David E.. ''Differential Equations and Boundary Value Problems'' 4th Edition. Upper Saddle River, NJ: Pearson, 2008</ref> is needed. For the first order equation, the integrating factor is defined as
<math> p = e^{\int F(x)} </math>
Multiplying both sides of the first order equation yields
<math> \frac{dy}{dx}e^{\int F(x)} + F(x)e^{\int F(x)}y = e^{\int F(x)}G(x) </math>
Note the derivative of
<math> e^{\int F(x)} </math>
is
<math> F(x)e^{\int F(x)} </math>
Hence the left hand side of the first order equation now looks like the [[product rule]] expansion for
<math> y </math> and <math> e^{\int F(x)} </math>
The equation can be rewritten
<math> \frac{d(e^{\int F(x)}y)}{dx} = e^{\int F(x)}G(x) </math>
Now we can integrate both sides, yielding the solution y:
<math> \int\frac{d(e^{\int F(x)}y)}{dx} = \int e^{\int F(x)}G(x) </math>
<math> e^{\int F(x)}y = \int e^{\int F(x)}G(x) + C </math>
<math> y = e^{-\int F(x)} \int e^{\int F(x)}G(x) + Ce^{-\int F(x)} </math>
where C is an arbitrary constant.
==References==
==References==