Solve the following problem, $u'(t)+p(t)u(t)=0,;;u(0)=0,$ $p(t)=begincases2& 0leq t< 1,\1 &tgeq 1endcases.$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Using Laplace transform, solve the following problem.
$$u'(t)+p(t)u(t)=0,;;u(0)=0,$$
$$p(t)=begincases2& 0leq t< 1,\1 &tgeq 1endcases.$$
Here's what I've done:
Taking the Laplace transform of both sides
$$L(u'(t))+L(p(t)u(t))=0,$$
$$sU(s)-u(0)+L(p(t)u(t))=0,$$
$$sU(s)+L(p(t)u(t))=0.$$
I'm stuck at this point. Please, how do I proceed?
calculus algebra-precalculus differential-equations derivatives laplace-transform
asked Aug 16 at 2:23
Mike
75215
add a comment |Â
up vote
2
down vote
favorite
Using Laplace transform, solve the following problem.
$$u'(t)+p(t)u(t)=0,;;u(0)=0,$$
$$p(t)=begincases2& 0leq t< 1,\1 &tgeq 1endcases.$$
Here's what I've done:
Taking the Laplace transform of both sides
$$L(u'(t))+L(p(t)u(t))=0,$$
$$sU(s)-u(0)+L(p(t)u(t))=0,$$
$$sU(s)+L(p(t)u(t))=0.$$
I'm stuck at this point. Please, how do I proceed?
calculus algebra-precalculus differential-equations derivatives laplace-transform
asked Aug 16 at 2:23
Mike
75215
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Using Laplace transform, solve the following problem.
$$u'(t)+p(t)u(t)=0,;;u(0)=0,$$
$$p(t)=begincases2& 0leq t< 1,\1 &tgeq 1endcases.$$
Here's what I've done:
Taking the Laplace transform of both sides
$$L(u'(t))+L(p(t)u(t))=0,$$
$$sU(s)-u(0)+L(p(t)u(t))=0,$$
$$sU(s)+L(p(t)u(t))=0.$$
I'm stuck at this point. Please, how do I proceed?
calculus algebra-precalculus differential-equations derivatives laplace-transform
asked Aug 16 at 2:23
Mike
75215
Using Laplace transform, solve the following problem.
$$u'(t)+p(t)u(t)=0,;;u(0)=0,$$
$$p(t)=begincases2& 0leq t< 1,\1 &tgeq 1endcases.$$
Here's what I've done:
Taking the Laplace transform of both sides
$$L(u'(t))+L(p(t)u(t))=0,$$
$$sU(s)-u(0)+L(p(t)u(t))=0,$$
$$sU(s)+L(p(t)u(t))=0.$$
I'm stuck at this point. Please, how do I proceed?
calculus algebra-precalculus differential-equations derivatives laplace-transform
asked Aug 16 at 2:23
Mike
75215
edited Aug 16 at 4:27
asked Aug 16 at 2:23
Mike
75215
asked Aug 16 at 2:23
Mike
75215
asked Aug 16 at 2:23
Mike
75215
75215
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
For $0< t< 1$, we have $u'(t)+2u(t)=0$, $u(0)=0$ for which there is trivial solution $u(t)=0$.
For $tge 1$, we have $u'(t)+u(t)=0$ for which the solution is $u(t)=Ae^-t$ where $A$ is arbitrary constant.
Added: As per David's comments, using the continuity of $u(t)$ at $t=1$ we get $A=0$. Hence the solution will be $u(t)=0$ for $tge 0$
answered Aug 16 at 3:51
Mathlover
3,6211021
But $u(t)$ is one function, not two separate functions. So the first part gives $u(1)=0$ which is an initial condition for the second part. And this gives $A=0$. So your solution is: $u(t)=0$ for all $tge0$.
– David
Aug 16 at 5:37
1
@David...Is it necessary that $u$ is continuous at $t=1$ as there are actually two ODE's defined in different domain.
– Mathlover
Aug 16 at 5:41
I would not say there are two DEs, but this is probably a matter of terminology so I won't argue. However it seems clear that the OP is asking about one function. The DE is given for all $tge0$ so we need a solution which is defined and differentiable, therefore also continuous, for all $tge0$.
– David
Aug 16 at 5:48
@David...I took account of the possibility that $u(t)$ may be piecewise continuous for $tge 0$.
– Mathlover
Aug 16 at 5:50
Well in that case the given DE is not satisfied at $t=1$.
– David
Aug 16 at 5:50
 |Â
show 2 more comments
up vote
1
down vote
$$int_0dfracd u(t)u(t)=int_0-p(t) dt=begincases2t& 0leq tleq 1,\t &tgeq 1endcases.$$
then $$u(t)=begincases0& tleq 0,\e^2t& 0leq tleq 1,\e^t &tgeq 1endcases.$$
answered Aug 16 at 3:30
Nosrati
20.6k41644
I guess it should be $e^-2t$ and $e^-t$
– Mike
Aug 16 at 4:49
yes they are. thanks.
– Nosrati
Aug 16 at 7:29
Please, can you teach me how to take inverse Laplace to get the results?
– Mike
Aug 16 at 7:30
Please, see: math.stackexchange.com/questions/2884434/…
– Mike
Aug 16 at 7:32
I didn't use laplace method.
– Nosrati
Aug 16 at 7:39
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For $0< t< 1$, we have $u'(t)+2u(t)=0$, $u(0)=0$ for which there is trivial solution $u(t)=0$.
For $tge 1$, we have $u'(t)+u(t)=0$ for which the solution is $u(t)=Ae^-t$ where $A$ is arbitrary constant.
Added: As per David's comments, using the continuity of $u(t)$ at $t=1$ we get $A=0$. Hence the solution will be $u(t)=0$ for $tge 0$
answered Aug 16 at 3:51
Mathlover
3,6211021
But $u(t)$ is one function, not two separate functions. So the first part gives $u(1)=0$ which is an initial condition for the second part. And this gives $A=0$. So your solution is: $u(t)=0$ for all $tge0$.
– David
Aug 16 at 5:37
1
@David...Is it necessary that $u$ is continuous at $t=1$ as there are actually two ODE's defined in different domain.
– Mathlover
Aug 16 at 5:41
I would not say there are two DEs, but this is probably a matter of terminology so I won't argue. However it seems clear that the OP is asking about one function. The DE is given for all $tge0$ so we need a solution which is defined and differentiable, therefore also continuous, for all $tge0$.
– David
Aug 16 at 5:48
@David...I took account of the possibility that $u(t)$ may be piecewise continuous for $tge 0$.
– Mathlover
Aug 16 at 5:50
Well in that case the given DE is not satisfied at $t=1$.
– David
Aug 16 at 5:50
 |Â
show 2 more comments
up vote
1
down vote
accepted
For $0< t< 1$, we have $u'(t)+2u(t)=0$, $u(0)=0$ for which there is trivial solution $u(t)=0$.
For $tge 1$, we have $u'(t)+u(t)=0$ for which the solution is $u(t)=Ae^-t$ where $A$ is arbitrary constant.
Added: As per David's comments, using the continuity of $u(t)$ at $t=1$ we get $A=0$. Hence the solution will be $u(t)=0$ for $tge 0$
answered Aug 16 at 3:51
Mathlover
3,6211021
But $u(t)$ is one function, not two separate functions. So the first part gives $u(1)=0$ which is an initial condition for the second part. And this gives $A=0$. So your solution is: $u(t)=0$ for all $tge0$.
– David
Aug 16 at 5:37
1
@David...Is it necessary that $u$ is continuous at $t=1$ as there are actually two ODE's defined in different domain.
– Mathlover
Aug 16 at 5:41
I would not say there are two DEs, but this is probably a matter of terminology so I won't argue. However it seems clear that the OP is asking about one function. The DE is given for all $tge0$ so we need a solution which is defined and differentiable, therefore also continuous, for all $tge0$.
– David
Aug 16 at 5:48
@David...I took account of the possibility that $u(t)$ may be piecewise continuous for $tge 0$.
– Mathlover
Aug 16 at 5:50
Well in that case the given DE is not satisfied at $t=1$.
– David
Aug 16 at 5:50
 |Â
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For $0< t< 1$, we have $u'(t)+2u(t)=0$, $u(0)=0$ for which there is trivial solution $u(t)=0$.
For $tge 1$, we have $u'(t)+u(t)=0$ for which the solution is $u(t)=Ae^-t$ where $A$ is arbitrary constant.
Added: As per David's comments, using the continuity of $u(t)$ at $t=1$ we get $A=0$. Hence the solution will be $u(t)=0$ for $tge 0$
answered Aug 16 at 3:51
Mathlover
3,6211021
For $0< t< 1$, we have $u'(t)+2u(t)=0$, $u(0)=0$ for which there is trivial solution $u(t)=0$.
For $tge 1$, we have $u'(t)+u(t)=0$ for which the solution is $u(t)=Ae^-t$ where $A$ is arbitrary constant.
Added: As per David's comments, using the continuity of $u(t)$ at $t=1$ we get $A=0$. Hence the solution will be $u(t)=0$ for $tge 0$
answered Aug 16 at 3:51
Mathlover
3,6211021
edited Aug 16 at 5:56
answered Aug 16 at 3:51
Mathlover
3,6211021
answered Aug 16 at 3:51
Mathlover
3,6211021
answered Aug 16 at 3:51
Mathlover
3,6211021
3,6211021
But $u(t)$ is one function, not two separate functions. So the first part gives $u(1)=0$ which is an initial condition for the second part. And this gives $A=0$. So your solution is: $u(t)=0$ for all $tge0$.
– David
Aug 16 at 5:37
1
@David...Is it necessary that $u$ is continuous at $t=1$ as there are actually two ODE's defined in different domain.
– Mathlover
Aug 16 at 5:41
I would not say there are two DEs, but this is probably a matter of terminology so I won't argue. However it seems clear that the OP is asking about one function. The DE is given for all $tge0$ so we need a solution which is defined and differentiable, therefore also continuous, for all $tge0$.
– David
Aug 16 at 5:48
@David...I took account of the possibility that $u(t)$ may be piecewise continuous for $tge 0$.
– Mathlover
Aug 16 at 5:50
Well in that case the given DE is not satisfied at $t=1$.
– David
Aug 16 at 5:50
 |Â
show 2 more comments
But $u(t)$ is one function, not two separate functions. So the first part gives $u(1)=0$ which is an initial condition for the second part. And this gives $A=0$. So your solution is: $u(t)=0$ for all $tge0$.
– David
Aug 16 at 5:37
1
@David...Is it necessary that $u$ is continuous at $t=1$ as there are actually two ODE's defined in different domain.
– Mathlover
Aug 16 at 5:41
I would not say there are two DEs, but this is probably a matter of terminology so I won't argue. However it seems clear that the OP is asking about one function. The DE is given for all $tge0$ so we need a solution which is defined and differentiable, therefore also continuous, for all $tge0$.
– David
Aug 16 at 5:48
@David...I took account of the possibility that $u(t)$ may be piecewise continuous for $tge 0$.
– Mathlover
Aug 16 at 5:50
Well in that case the given DE is not satisfied at $t=1$.
– David
Aug 16 at 5:50
But $u(t)$ is one function, not two separate functions. So the first part gives $u(1)=0$ which is an initial condition for the second part. And this gives $A=0$. So your solution is: $u(t)=0$ for all $tge0$.
– David
Aug 16 at 5:37
But $u(t)$ is one function, not two separate functions. So the first part gives $u(1)=0$ which is an initial condition for the second part. And this gives $A=0$. So your solution is: $u(t)=0$ for all $tge0$.
– David
Aug 16 at 5:37
1
1
@David...Is it necessary that $u$ is continuous at $t=1$ as there are actually two ODE's defined in different domain.
– Mathlover
Aug 16 at 5:41
@David...Is it necessary that $u$ is continuous at $t=1$ as there are actually two ODE's defined in different domain.
– Mathlover
Aug 16 at 5:41
I would not say there are two DEs, but this is probably a matter of terminology so I won't argue. However it seems clear that the OP is asking about one function. The DE is given for all $tge0$ so we need a solution which is defined and differentiable, therefore also continuous, for all $tge0$.
– David
Aug 16 at 5:48
I would not say there are two DEs, but this is probably a matter of terminology so I won't argue. However it seems clear that the OP is asking about one function. The DE is given for all $tge0$ so we need a solution which is defined and differentiable, therefore also continuous, for all $tge0$.
– David
Aug 16 at 5:48
@David...I took account of the possibility that $u(t)$ may be piecewise continuous for $tge 0$.
– Mathlover
Aug 16 at 5:50
@David...I took account of the possibility that $u(t)$ may be piecewise continuous for $tge 0$.
– Mathlover
Aug 16 at 5:50
Well in that case the given DE is not satisfied at $t=1$.
– David
Aug 16 at 5:50
Well in that case the given DE is not satisfied at $t=1$.
– David
Aug 16 at 5:50
 |Â
show 2 more comments
up vote
1
down vote
$$int_0dfracd u(t)u(t)=int_0-p(t) dt=begincases2t& 0leq tleq 1,\t &tgeq 1endcases.$$
then $$u(t)=begincases0& tleq 0,\e^2t& 0leq tleq 1,\e^t &tgeq 1endcases.$$
answered Aug 16 at 3:30
Nosrati
20.6k41644
I guess it should be $e^-2t$ and $e^-t$
– Mike
Aug 16 at 4:49
yes they are. thanks.
– Nosrati
Aug 16 at 7:29
Please, can you teach me how to take inverse Laplace to get the results?
– Mike
Aug 16 at 7:30
Please, see: math.stackexchange.com/questions/2884434/…
– Mike
Aug 16 at 7:32
I didn't use laplace method.
– Nosrati
Aug 16 at 7:39
add a comment |Â
up vote
1
down vote
$$int_0dfracd u(t)u(t)=int_0-p(t) dt=begincases2t& 0leq tleq 1,\t &tgeq 1endcases.$$
then $$u(t)=begincases0& tleq 0,\e^2t& 0leq tleq 1,\e^t &tgeq 1endcases.$$
answered Aug 16 at 3:30
Nosrati
20.6k41644
I guess it should be $e^-2t$ and $e^-t$
– Mike
Aug 16 at 4:49
yes they are. thanks.
– Nosrati
Aug 16 at 7:29
Please, can you teach me how to take inverse Laplace to get the results?
– Mike
Aug 16 at 7:30
Please, see: math.stackexchange.com/questions/2884434/…
– Mike
Aug 16 at 7:32
I didn't use laplace method.
– Nosrati
Aug 16 at 7:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$int_0dfracd u(t)u(t)=int_0-p(t) dt=begincases2t& 0leq tleq 1,\t &tgeq 1endcases.$$
then $$u(t)=begincases0& tleq 0,\e^2t& 0leq tleq 1,\e^t &tgeq 1endcases.$$
answered Aug 16 at 3:30
Nosrati
20.6k41644
$$int_0dfracd u(t)u(t)=int_0-p(t) dt=begincases2t& 0leq tleq 1,\t &tgeq 1endcases.$$
then $$u(t)=begincases0& tleq 0,\e^2t& 0leq tleq 1,\e^t &tgeq 1endcases.$$
answered Aug 16 at 3:30
Nosrati
20.6k41644
answered Aug 16 at 3:30
Nosrati
20.6k41644
answered Aug 16 at 3:30
Nosrati
20.6k41644
answered Aug 16 at 3:30
Nosrati
20.6k41644
20.6k41644
I guess it should be $e^-2t$ and $e^-t$
– Mike
Aug 16 at 4:49
yes they are. thanks.
– Nosrati
Aug 16 at 7:29
Please, can you teach me how to take inverse Laplace to get the results?
– Mike
Aug 16 at 7:30
Please, see: math.stackexchange.com/questions/2884434/…
– Mike
Aug 16 at 7:32
I didn't use laplace method.
– Nosrati
Aug 16 at 7:39
add a comment |Â
I guess it should be $e^-2t$ and $e^-t$
– Mike
Aug 16 at 4:49
yes they are. thanks.
– Nosrati
Aug 16 at 7:29
Please, can you teach me how to take inverse Laplace to get the results?
– Mike
Aug 16 at 7:30
Please, see: math.stackexchange.com/questions/2884434/…
– Mike
Aug 16 at 7:32
I didn't use laplace method.
– Nosrati
Aug 16 at 7:39
I guess it should be $e^-2t$ and $e^-t$
– Mike
Aug 16 at 4:49
I guess it should be $e^-2t$ and $e^-t$
– Mike
Aug 16 at 4:49
yes they are. thanks.
– Nosrati
Aug 16 at 7:29
yes they are. thanks.
– Nosrati
Aug 16 at 7:29
Please, can you teach me how to take inverse Laplace to get the results?
– Mike
Aug 16 at 7:30
Please, can you teach me how to take inverse Laplace to get the results?
– Mike
Aug 16 at 7:30
Please, see: math.stackexchange.com/questions/2884434/…
– Mike
Aug 16 at 7:32
Please, see: math.stackexchange.com/questions/2884434/…
– Mike
Aug 16 at 7:32
I didn't use laplace method.
– Nosrati
Aug 16 at 7:39
I didn't use laplace method.
– Nosrati
Aug 16 at 7:39
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2884318%2fsolve-the-following-problem-utptut-0-u0-0-pt-begincases2%23new-answer', 'question_page');
);
Post as a guest
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
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
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