- #1

DryRun

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**Homework Statement**

[tex]y''+4y'+3y=\sinh x, \, y(0)=y'(0)=0[/tex]

**The attempt at a solution**

The auxiliary equation is: [itex]m^2+4m+3=0[/itex]

m= -3 and -1

The complementary function is: [itex]y_c=Ae^{-3x}+Be^{-x}[/itex]

I need to find the particular integral next. But i don't know what is the standard inverse D operator result in this case, but i'm going to deduce it from what information i do know.

[itex]L(D^2)\sinh ax=L(a^2)\sinh ax[/itex]

From this, i will deduce (maybe i'm wrong) the principle [itex]\frac{1}{L(D^2)}\sinh ax=\frac{1}{L(a^2)}\sinh ax[/itex], where [itex]L(a^2)\not =0[/itex]

[itex]L(D)=D^2+4D+3[/itex]

a=1, k=1

[itex]y_p=\frac{1}{1+4D+3}\sinh x=\frac{1}{4D+4}\sinh x[/itex]

On multiplying numerator and denominator by (4D-4), [itex]\frac{4D-4}{16D^2-16}\sinh x[/itex]

And then i applied the same principle again: [itex]\frac{4D-4}{0}\sinh x[/itex]

And then i got stuck, since the denominator is zero. I must have done something wrong somewhere but i have no idea where.

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