Laplace Transform, inverse Laplace transform differential equation
up vote
0
down vote
favorite
how would you find the solution x(t) of $frac{dx}{dt} = rx + 1000sum_{i=1}^infty delta(t-i)$ for $x(0)=0$ using laplace transformation?
Am I heading in the right direction as I am stucked?
$sX(s)-rX(s) = 1000sum_{i=1}^infty e^{-is}$
$X(s) = frac{1000e^{-s}}{(1-e^{-s})(s-r)}$
differential-equations laplace-transform exponential-sum
asked Dec 1 at 12:33
G.L.
204
add a comment |
up vote
0
down vote
favorite
how would you find the solution x(t) of $frac{dx}{dt} = rx + 1000sum_{i=1}^infty delta(t-i)$ for $x(0)=0$ using laplace transformation?
Am I heading in the right direction as I am stucked?
$sX(s)-rX(s) = 1000sum_{i=1}^infty e^{-is}$
$X(s) = frac{1000e^{-s}}{(1-e^{-s})(s-r)}$
differential-equations laplace-transform exponential-sum
asked Dec 1 at 12:33
G.L.
204
Looks right. However how to get the solution from that is unclear. Without Laplace, you would use $$(e^{-rt}x(t))'=1000sum_{i=1}^infty e^{-ir}δ(t−i) implies x(t)=1000sum_{i=1}^infty e^{(t-i)r}u(t-i),$$ and there is no way to further simplify the step discontinuities in that function, that is, to get rid of the infinite sum.
– LutzL
Dec 1 at 13:19
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
how would you find the solution x(t) of $frac{dx}{dt} = rx + 1000sum_{i=1}^infty delta(t-i)$ for $x(0)=0$ using laplace transformation?
Am I heading in the right direction as I am stucked?
$sX(s)-rX(s) = 1000sum_{i=1}^infty e^{-is}$
$X(s) = frac{1000e^{-s}}{(1-e^{-s})(s-r)}$
differential-equations laplace-transform exponential-sum
asked Dec 1 at 12:33
G.L.
204
how would you find the solution x(t) of $frac{dx}{dt} = rx + 1000sum_{i=1}^infty delta(t-i)$ for $x(0)=0$ using laplace transformation?
Am I heading in the right direction as I am stucked?
$sX(s)-rX(s) = 1000sum_{i=1}^infty e^{-is}$
$X(s) = frac{1000e^{-s}}{(1-e^{-s})(s-r)}$
differential-equations laplace-transform exponential-sum
differential-equations laplace-transform exponential-sum
asked Dec 1 at 12:33
G.L.
204
asked Dec 1 at 12:33
G.L.
204
asked Dec 1 at 12:33
G.L.
204
asked Dec 1 at 12:33
G.L.
204
asked Dec 1 at 12:33
G.L.
204
204
Looks right. However how to get the solution from that is unclear. Without Laplace, you would use $$(e^{-rt}x(t))'=1000sum_{i=1}^infty e^{-ir}δ(t−i) implies x(t)=1000sum_{i=1}^infty e^{(t-i)r}u(t-i),$$ and there is no way to further simplify the step discontinuities in that function, that is, to get rid of the infinite sum.
– LutzL
Dec 1 at 13:19
add a comment |
Looks right. However how to get the solution from that is unclear. Without Laplace, you would use $$(e^{-rt}x(t))'=1000sum_{i=1}^infty e^{-ir}δ(t−i) implies x(t)=1000sum_{i=1}^infty e^{(t-i)r}u(t-i),$$ and there is no way to further simplify the step discontinuities in that function, that is, to get rid of the infinite sum.
– LutzL
Dec 1 at 13:19
Looks right. However how to get the solution from that is unclear. Without Laplace, you would use $$(e^{-rt}x(t))'=1000sum_{i=1}^infty e^{-ir}δ(t−i) implies x(t)=1000sum_{i=1}^infty e^{(t-i)r}u(t-i),$$ and there is no way to further simplify the step discontinuities in that function, that is, to get rid of the infinite sum.
– LutzL
Dec 1 at 13:19
Looks right. However how to get the solution from that is unclear. Without Laplace, you would use $$(e^{-rt}x(t))'=1000sum_{i=1}^infty e^{-ir}δ(t−i) implies x(t)=1000sum_{i=1}^infty e^{(t-i)r}u(t-i),$$ and there is no way to further simplify the step discontinuities in that function, that is, to get rid of the infinite sum.
– LutzL
Dec 1 at 13:19
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021278%2flaplace-transform-inverse-laplace-transform-differential-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Looks right. However how to get the solution from that is unclear. Without Laplace, you would use $$(e^{-rt}x(t))'=1000sum_{i=1}^infty e^{-ir}δ(t−i) implies x(t)=1000sum_{i=1}^infty e^{(t-i)r}u(t-i),$$ and there is no way to further simplify the step discontinuities in that function, that is, to get rid of the infinite sum.
– LutzL
Dec 1 at 13:19