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FirstOrderLinearDE#2 - F ir s t-O r d e r L in e a r D iffe...

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Page 1 of 12 First-Order Linear Differential Equations #2 Generating New Differential Equations using a Multiplier Function In this section we show how to obtain new first-order differential equations by transforming the solutions of a given first-order DE by multiplying them by a given function m ( t ). One application of this theory is that the new differential equation is easier to solve if the multiplier is chosen judiciously. Example : In a previous section, we saw the solutions of the 1 st -order differential equation: dx dt = 2 t DE1 are the family of vertically-shifted parabolas shown to the right and given explicitly below. x c ( t ) = t 2 + c The critical curve is the vertical axis t = 0 . See the dotted black curve. If we multiply each of the above solutions by a multiplier function such as m ( t ) = e t we obtain the new family of functions z c ( t ) shown to the right and defined below by: z c ( t ) m ( t ) Multiplier Function x c ( t ) = e t t 2 + c ( ) The new critical curve is z = e t 2 t . See the black curve. Does this new family of functions also satisfy a differential equation? Yes. We will verify that this new family of curves satisfies the new differential equation: dz dt + z = e t 2 t DE2 Critical curve shown in black. Using the product rule, the derivative of z is seen to be: dz dt = e t t 2 + c ( ) z ( t )   + e t 2 t Recognizing the term z ( t ) on the RHS, we bring it to the left and obtain: dz dt + z = e t 2 t In this case, the new differential equation DE2 appears harder than the first one since DE1 has the form of a perfect derivative: dx dt = f ( t ) . But we could easily run the above process in reverse. If we started with DE2, it can be converted to the simpler DE1 by multiplying by the reciprocal of the multiplier. This is the idea behind Leibniz’s multiplier method. Another interesting question is how this process works in general. How can we find the new differential equation that results when the solutions to a given differential equation are multiplied by a multiplier function m ( t )? The answer is revealed in the following theorem.
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Page 2 of 12 Multiplier Theorem A. Multiplier functions transform 1 st -order linear DEs to new 1 st -order linear DEs. Suppose the function y ( t ) is a solution to the first-order linear differential equation: dy dt + p ( t ) y ( t ) = g ( t ) DE1 Next multiply each such solution by the multiplier function m ( t ), obtaining the product function: z ( t ) = m ( t ) y ( t ) Then the product function z ( t ) is a solution to the new first-order linear differential equation: dz dt + p ( t ) m '( t ) m ( t ) P ( t )   z ( t ) = m ( t ) g ( t ) G ( t )   DE2 B. You can already see one potential application. The coefficient of z ( t ) can be eliminated if we slyly choose the multiplier function m ( t ) so that p ( t ) m '( t ) m ( t ) = 0 . This is exactly the idea behind Leibniz’s method. Rearranging the above equation we find dm m = p ( t ) dt for which one solution is Leibniz’s magic factor m ( t ) = e p ( t ) dt .
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