[itex]y''-3y'+2y=\mathrm{e}^x +1[/itex](adsbygoogle = window.adsbygoogle || []).push({});

Find [itex]y[/itex].

Going through the normal procedure we get the complementary function

[itex]y = A\mathrm{e}^x + B\mathrm{e}^{2x}[/itex].

Using the trial function

[itex]y = Cx\mathrm{e}^x + D[/itex]

(because of overlap with the complementary function)

leads to

[itex]-C\mathrm{e}^x + 2D = \mathrm{e}^x+1[/itex]

Does this mean

[itex]C=-1,D=1/2[/itex]??

If so, the general soln is [itex]y = A\mathrm{e}^x + B\mathrm{e}^{2x} -x\mathrm{e}^x + 1/2[/itex].

According to Mathematica, however, the solution should be of the form

[itex]y = A\mathrm{e}^x + B\mathrm{e}^{2x} + 1/2(1-2\mathrm{e}^x -2\mathrm{e}^xx)[/itex].

Am I using the wrong trial function or what?

Thanks in advance.

James

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# Homework Help: Trial function for ODE (Exam tomorrow)

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