- #1
jdstokes
- 523
- 1
[itex]y''-3y'+2y=\mathrm{e}^x +1[/itex]
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
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