Problem on Inverse Laplace with Unit Step

[purduegrad]

I have the equation (t-3)u2(t) - (t-2)u3(t) and they want the inverse laplace transformation of this. So basicially I drew out the unit step graph. And i had 0<t<2 --->0; 2<t<3 ---> t-3 and t>3 ----> -1. So then I just did wrote out the laplace transformations and I got

(s-3)[(-e^-3s)/s + (e^-2s)/s] - (e^-3s)/s


There is the part where I'm stuck at...I'm not sure if I am right regarding this part, but some assistance is definitely needed. Thanks guys!

Jason

 

[Tide]

I don't understand your notation. What are u2 and u3?

Also, you said they are asking for the inverse Laplace transform when you're starting in the time domain. Since you're ending up in the s-domain I assume you meant the forward transform.

 

[M98Ranger]

Hi Jason,

Based on your explanation, it seems like you have the correct approach for finding the inverse Laplace transformation of (t-3)u2(t) - (t-2)u3(t). As you mentioned, you first need to determine the Laplace transformation of each piece of the unit step function.

For 0<t<2, the Laplace transformation is 0. For 2<t<3, the Laplace transformation is (t-3)/s. And for t>3, the Laplace transformation is -1/s.

Next, you can use the linearity property of the Laplace transformation to combine these individual transformations. So your final inverse Laplace transformation will be:

L^-1[(t-3)u2(t) - (t-2)u3(t)] = L^-1[0 + (t-3)/s - 1/s] = L^-1[(t-2)/s - 1/s] = L^-1[(t-3)/s] = e^3t

I hope this helps clarify things for you. Let me know if you have any other questions. Good luck!