Second-order differential equation and conditions

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SUMMARY

The discussion focuses on solving a second-order differential equation, specifically addressing the limit conditions in part d), ii). The equation presented is ##y=c_1e^{2x}+c_2e^{-3x}-e^{-x}##. The key issue raised is the misinterpretation of limits leading to the incorrect conclusion of ##0=\infty##. The correct approach involves evaluating the limits separately to determine the values of the constants that satisfy the equation as x approaches infinity.

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Taylor_1989
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Homework Statement


Hy guys I am have a problem with the last part of this question. part d), ii) I get the general formal which I have displayed below, but what I done understand is if I take the limits as show in ii) I get ##0=\
\infty## which obviously I am doing something wrong. Have I misinterpreted what they are say as y tends to 0 as x tends. Could anyone give me some advice it would be much appreciated, thanks in advance.
upload_2017-3-30_16-35-20.png

Homework Equations

The Attempt at a Solution


##y=c_1e^{2x}+c_2e^{-3x}-e^{-x}##
 

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Taylor_1989 said:

Homework Statement


Hy guys I am have a problem with the last part of this question. part d), ii) I get the general formal which I have displayed below, but what I done understand is if I take the limits as show in ii) I get ##0=\
\infty## which obviously I am doing something wrong. Have I misinterpreted what they are say as y tends to 0 as x tends. Could anyone give me some advice it would be much appreciated, thanks in advance.
View attachment 115277

Homework Equations

The Attempt at a Solution


##y=c_1e^{2x}+c_2e^{-3x}-e^{-x}##
What equation did you get for question d) part i)? Part ii) is essentially saying that ##\lim_{x \to \infty} c_1e^{2x} + c_2e^{-3x} - e^{-x} = 0##. What are the values of the two constants so that this can happen?
 
@Mark44 I see now thank you as soo as u put in that form and do each lim seperatly I saw what was going on. Much apprectied.
 

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