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Jon D.

Differential Equations

1 week, 3 days ago

Okay, so we got our function e to the negative 3 s over s plus 2 squared multiplied by s minus 1 point now: this 1 can be rewritten as e to the negative 3 s over s plus 2 squared multiplied by 1 over s, minus 1 point And now this guy is what okay, so first of all, let me define the function f of t equals t multiplied by e to the negative 2 t point, then what do we have? Well, we have that e to the negative to 3 s over s plus 2. Squared is the laplace transform of f of t minus 3 multiplied by gamma orb t minus 3, where gamma of s is the function defined by 1, if s is greater than 00, if s is less than or equal to, 0 multiplied by the lap and d Is 1 here is the la plastron storm of e? To the the point now, let's use the volution theorem to compute the inverse laplace transform of uato thanks to the lion. Theorem is equal to any integral from 0 to 3 of f of t minus 3 minus 2 gamma of t minus 3 minus no multiplied by e to the new interne okay. Now here we are going to use a change of variable to compute this integral. Here we are going to set as equals t minus 3 minus not in such a way. The s is equal to negative now, and so here we're going to have what well, when the t is equal to 0 point. Okay, so when t is equal to 0, here we get negative 3 negative when okay, actually, when i know, is equal to 0 when no is equal to 0. Here we get t minus 3 and when no is equal to t okay here when t is equal to no here we get negative 3 point. So here we got an integral of f of s, multiplied by gamma of s, and here we got t minus 3 minus sindon. So now we can revise this guy as any integral from negative 3 to t minus 3 of f of s multiplied by gamma of s. Okay and actually we can take e to the t to the t- minus 3 here, okay, so here we get just e to the negative s in doin. Now what do we know here? We know that this guy is okay. 0 point! Okay! So let's see this, guy is 0 if t is less than or equal to 3 and is e to the t: minus 3 multiplied by an integral from 0 to t minus 3 of f of s e to the negative. As in d s point: okay! Now we know that f of s is just as multiplied by e to the negative 2 s. So here let me rewrite it here. We are going to have okay again, 0 for the less than or equal to 3, and here we got e to the t. Minus 3 multiplied by an integral of s e to the negative 3 s in the s from 0 to t minus 3 point so now we just need to compute an anti derivative of this 1 and well. Okay, i'm going to write it here. So s e to the negative 3 s and s here we are going to use integration by parts. So f here grim here and here we got negative s over 3, o okay. So this 1 is f multiplied by j e to the negative 3 s. Minus g f, prime, so plus 1 over 3 and any integral of e to the negative 3 s. So this 1 is going to be negative s over 3 e to the negative 3 s minus 1, over 9 e to the negative 3 s. Okay, perfect! So okmaybe, let me rewrite it! This 1 is the same thing as negative as over 3 plus 1, over 9 e to the negative 3 s which must be evaluated between between between t minus 3 and 0 point okay. So, let's compute this evaluation, so negative s over 3 plus here it's like this okay, so this 1 is going to be. Actually maybe it's better if we were at it like this. This 1 is s over 3 plus 1 over 9 multiplied by e to the negative 3 s evaluated between 0 and minus 3 point. So here we get 1 over 9 minus okay, so minus 3 over 3, okay, minus 1, over 9 minus 1. Actually, here it's going to be like this. Okay, and here we got e to the negative 3 multiplied by t minus 3 point. Okay. So at this point we are done because we have that the inverse la plus transform- or u is equal to 0. If t is less than or equal to 3 and is given by we computed this integral here so here we are going to have multiplied by t minus 3, like this 10 e to the t, minus 3 multiplied by 19 minus t minus 3 over 3 plus 1, over 9 e to the negative 3 t minus 34 t greater than or equal to 3 point, and so we are.

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