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Find general solution of differential eqn

  • Thread starter DryRun
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  • #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.
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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[itex]Sinh x= (e^x- e^{-x})/2[/itex] so that your solution to the homogeneous equation, [itex]e^{-x}[/itex] is part of the particular solution. Try a particular solution of the form [itex]Ae^x+ Bxe^{-x}[/itex].
 
  • #3
DryRun
Gold Member
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Unfortunately, i don't understand the method. I need a formula that i can always use whenever i see sinhx but it's unfortunately not listed in my notes.

For example, in my notes, [itex]\frac{1}{L(D)}ke^{ax}[/itex] gives the generic answer: [itex]\frac{1}{L(a)}ke^{ax}[/itex]

And also, in my notes, [itex]\frac{1}{L(D)}ksin(ax+b)[/itex] gives the generic answer: [itex]\frac{1}{L(-a^2)}ksin(ax+b)[/itex]

But there is nothing about [itex]\sinh x[/itex], so i deduced it myself from the above formula:
[itex]\frac{1}{L(D)}ksinh(ax+b)[/itex] gives the generic answer: [itex]\frac{1}{L(a^2)}ksinh(ax+b)[/itex] but somehow it's wrong as it doesn't solve the problem in post #1.
 
Last edited:
  • #4
DryRun
Gold Member
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I finally got it! HallsofIvy, i'm sorry for dismissing your suggestion without looking deeper into it. I converted the R.H.S. to [itex]\frac{e^x-e^{-x}}{2}[/itex] which is the equivalent of [itex]\sinh x[/itex] and the correct answer revealed itself. Thank you for your help.
 

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