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

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

is the given answer wrong ?

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

Originally Posted by xl5899
http://i.imgur.com/JhexTZe.jpg

image here

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 ?