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can someone kindly help me with these few questions? :)

Find $L{e^tf(t)}$ in terms of $f*(s)$ and state a range of $s$ which this is defined.

I couldn't figure this out. You use the definition but i then get $e^t(1-s)f(t)$ and then i don't know what to do....

Using the Convolution Theorem, find the function f(t) satisfying the equation

$f(t) = \int_0^t e^u f(t-u)\mathrm du + e^t$

I know how to take the LT of both sides. I'm not sure how to figure out the LT of the integral though. Ive for f(s) = something + 1/(s-1) at the moment.

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    user4645: So did you intend $f(t) = \int_0^t {ue^u f(t - u)du} + e^t $?2011-05-23

2 Answers 2

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Here is a detailed solution to the modified -- substantially more challenging -- problem (see the OP's comments below the previous answer; in particular, it is stated there that this is not homework).

To find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation $ f(t) = \int_0^t {u e^u f(t - u)\,du} + e^t, \;\; t \geq 0, $ begin, as in the previous answer, by writing $ \hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1, $ where this time $\hat \varphi$ is the Laplace transform of the convolution $ (f*te^t )(t) = \int_0^t {ue^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)(t-u)e^{t - u}\,du}\bigg). $ By the convolution theorem, $ \hat \varphi(s) = \hat f(s) \frac{1}{{(s - 1)^2 }}. $ It is worth noting that the term $1/(s-1)^2$ can be derived as follows, recalling that an exponential random variable with density function $\lambda e^{-\lambda t}$, $t \geq 0$, has mean equal to $1/\lambda$ (here \lambda = s -1 > 0): $ \int_0^\infty {e^{ - st} te^t \,dt} = \frac{1}{{s - 1}}\int_0^\infty {t(s - 1)e^{ - (s - 1)t} \,dt} = \frac{1}{{(s - 1)^2 }}. $ Solving for $\hat f(s)$ (using the above expression for $\hat \varphi(s)$) gives $ \hat f(s) = \frac{{s - 1}}{{s^2 - 2s}} = \frac{{(s - 2) + 1}}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{s}\frac{1}{{s - 2}}. $ Assuming that $s > 2$, it follows by inversion (and the Convolution Theorem) that $ f(t) = 1 + (1 * e^{2t})(t),\;\; t \geq 0. $ (Indeed, note that $\int_0^\infty {e^{ - st} 1\,dt} = \frac{1}{s}$ and $\int_0^\infty {e^{ - st} e^{2t} \,dt} = \frac{1}{{s - 2}}$.) Finally, from $ (1 * e^{2t})(t) = \int_0^t {e^{2u} 1\,du} \,\bigg( = \int_0^t {1e^{2(t - u)} \,du} \bigg) , $ it follows that $ f(t) = 1 + \frac{{e^{2t} - 1}}{2} = \frac{{e^{2t} + 1}}{2},\;\; t \geq 0. $ Indeed, this $f$ satisfies the original equation; that is, as one can easily verify, it holds $ \frac{{e^{2t} + 1}}{2} = \int_0^t {ue^u \frac{{e^{2(t - u)} + 1}}{2}\,du} + e^t . $

EDIT (in response to the OP's comment below). While inverting $1/s$ gives $1$ and inverting $1/(s-2)$ gives $e^{2t}$, inverting $1/(s(s-2))$ does not give the product of $e^{2t}$ and $1$; rather, by the Convolution Theorem, it gives the convolution of $e^{2t}$ and $1$. Since $(1 * e^{2t})(t) = \frac{{e^{2t} - 1}}{2}\,( = \int_0^t {e^{2u} \,du} )$, $ f(t) = 1 + \frac{{e^{2t} - 1}}{2} = \frac{{e{}^{2t} + 1}}{2} $ (which was verified by substitution into the original equation). However, as the OP observed, the solution can be obtained more elementarily by splitting $\hat f(s)$ into partial fractions. Specifically, $ \hat f(s) = \frac{{s - 1}}{{s^2 - 2s}} = \frac{1}{{2s}} + \frac{1}{{2(s - 2)}}, $ from which it follows, by inversion, that $ f(t) = \frac{1}{2} + \frac{1}{2}e^{2t} = \frac{{e^{2t} + 1}}{2}. $

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    okay -completely understood and now im getting most of the Laplace Transformation questions correct!!! Thanks so much. Its really helped!2011-05-25
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The solution to the first question follows easily from the definition: Can you bring $\int_0^\infty {e^{ - st} e^t f(t)\,dt}$ into the form \int_0^\infty {e^{ - s't} f(t) \,dt}? (Consider the hint you were already given above.)

The solution to the second question follows easily from the Convolution Theorem. You want to find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation $ f(t) = \int_0^t {e^u f(t - u)\,du} + e^t, \;\; t \geq 0. $ Your approach is right. Taking Laplace transform on both sides gives $ \hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1, $ where $\hat \varphi$ is the Laplace transform of the convolution $ (f*e^t )(t) = \int_0^t {e^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)e^{t - u}\,du}\bigg). $ (Note that the upper limit of the integral is $t$ in order to satisfy $t-u \geq 0$.) The Convolution Theorem states that the Laplace transform of a convolution is the product of the Laplace transforms of the individual functions. Hence, $ \hat \varphi(s) = \hat f(s) \frac{1}{{s - 1}}. $ Now you can solve for $\hat f(s)$, and, in turn by inversion (assuming that s > 2), find $f(t)$.

Note: After you find $f(t)$, verify that it satisfies the original equation. This is very easily done; indeed $f(t)$ is only slightly different from $e^t$ (in form).

EDIT (completing the solution; see Remark below):

Solving for $\hat f(s)$ gives $ \hat f(s) = \frac{1}{{s - 2}}. $ Assuming that $s > 2$, it follows by inversion that $f(t)=e^{2t}$. Indeed, this $f$ satisfies the original equation; that is, $ e^{2t} = \int_0^t {e^u e^{2(t - u)} \,du} + e^t ,\;\; t \geq 0. $

Remark. In view of the OP's comments below, it appears that the factor $u$ was forgotten in the convolution term of the original equation. A solution to the modified problem has now been posted.

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    Anyway, see the new answer, corresponding to the modified version (with $u$ added).2011-05-23