is the given answer wrong ?

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- Mar 10th 2016, 05:05 AM #1

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- Mar 10th 2016, 05:07 AM #2

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- Mar 10th 2016, 06:22 AM #3

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## Re: (d2x/dt2) +4(dx/dt) +3x= e^-3x

http://i.imgur.com/JhexTZe.jpg

image here

- Mar 10th 2016, 07:45 AM #4

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## Re: (d2x/dt2) +4(dx/dt) +3x= e^-3x

It's not that difficult to type the problem in rather than asking people you want to help you to open and image!

The problem you have is to show that x(t)= (1/2)(1- t)e^(-3t) satisfies the differential equation d^2 x/d^2+ 4dy/dx+ 3x= e^(-3t) with initial conditions x(0)= 1/2, x'(0)= -2.

First x(0)= (1/2)(1- 0)e^0= (1/2)(1)(1) so that the first condition, x(0)= 1/2, is satisfies.

Second, x'(t)= (1/2)(-1)e^(-3t)- (3/2)(1- t)e^(-3t)= (-2+ (3/2)t)e^(-3t) so x'(0)=(-2+ 0)e^0= -2 so the second condition, x'(0)= -2, is satisfied.

Finally, x''(t)= (3/2)e^(-3t)- (3)(-2+ (3/2)t)e^(-3t)= (15/2- (9/2)t)e^(3t) so that

x''+ 4x'+ 3x= (15/2- (9/2)t)e^(3t)+ 4(-2+ (3/2)t)e(-3t))+ 3(1/2)(1- t)e^(-3t)= [(15/2- 9+ 3/2)- (9/2+ 6- 3/2)t] e^(-3t)= -9 t e^(-3t) NOT e^(-3t).

- Mar 10th 2016, 03:02 PM #5

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- Mar 10th 2016, 03:28 PM #6

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