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

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

is the given answer wrong ?

4. ## 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).

5. ## Re: (d2x/dt2) +4(dx/dt) +3x= e^-3x Originally Posted by HallsofIvy 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).
if i wanna follow the original question , is my answer correct ?

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

i found the answer already , thanks for the help

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