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A problem with solutionby zoom1
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#1
Mar312, 06:05 PM

P: 23

Have an equation;
d^{2}y(t)/dt^{2} + 5d^{2}y(t)/dt^{2} + 4y(t) = 2e^{2t} Solved the complementary(homogenous) part and the function and got the roots of 1 and 4 so the y_{h}(t) is A_{1}.e^{4t} + A_{2}.e^{t} Forcing function is 2.e^{2t} so y_{particular}(t) is A.e^{2t} Am I right here ? Or am I supposed to use Ate^{2t} Well, if I use the first one, the resultant function doesn't give me the 2.e^{2t} when I put it into the differential equation, so there is something wrong obviously. However F(t) or one of its derivatives are not identical to terms in the homogenous solution, so I think I have to use the first option, which is A.e^{2t} After proceeding I ended up with y_{p}(t) = 1/3.e^{2t} Initial values are y(0) = 0 and y^{(1)}(0) = 0 so, K_{1} = 1/9 and K_{2} = 2/9 Still couldn't find where I am wrong Appreciate if you help me. 


#2
Mar312, 07:39 PM

PF Gold
P: 79




#3
Mar412, 04:36 AM

P: 23




#4
Mar412, 04:42 AM

Emeritus
Sci Advisor
PF Gold
P: 4,500

A problem with solution
You should double check your calculation for Ae^{2t}



#5
Mar412, 05:31 AM

P: 23




#6
Mar412, 06:37 PM

PF Gold
P: 79

I got a particular solution of y[itex]_{p}[/itex] = e[itex]^{2t}[/itex]. See if you get the same.



#7
Mar412, 08:20 PM

P: 2

I get [itex]A_1=\frac{1}{3}[/itex] and [itex]A_2=\frac{3}{2}[/itex]. The particular solution given above is correct.
Doublecheck your calculations. 


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