Mixing
A multiple choice question concerning first order linear differential equations
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Consider the initial value problem $$y'(t)=f(t)y(t)$$ with $y(0)=1$ where $f:mathbbRrightarrow mathbbR$ is a continuous function.
Then this initial value problem has
(1) Infinitely many solutions for some $f$
(2)a unique solution in $mathbbR$
(3) no solution in $mathbbR$ for some $f$
(4) a solution in an interval containing $0$, but not on $mathbbR$ for some $f$
My efforts
$$fracdy(t)y(t) = f(t)dt$$
$$log y(t) = int_0^tf(t)dt +C$$
Now at $t=0$ $y=1$ so we get
$$0=0+C$$
So $$y(t)=exp(int_0^tf(t)dt)$$
So either $2$ is true or $4$ is true? I am not able to go further from here. What logic should I use now
Edit:
Since $int_0^t f$ is continuous and $e^x$ is continuous and composition of continuous function is continuous so $2$ is true. Is this a correct way of thinking?
differential-equations initial-value-problems
asked 4 hours ago
StammeringMathematician
1,012217
add a comment |Â
up vote
4
down vote
favorite
Consider the initial value problem $$y'(t)=f(t)y(t)$$ with $y(0)=1$ where $f:mathbbRrightarrow mathbbR$ is a continuous function.
Then this initial value problem has
(1) Infinitely many solutions for some $f$
(2)a unique solution in $mathbbR$
(3) no solution in $mathbbR$ for some $f$
(4) a solution in an interval containing $0$, but not on $mathbbR$ for some $f$
My efforts
$$fracdy(t)y(t) = f(t)dt$$
$$log y(t) = int_0^tf(t)dt +C$$
Now at $t=0$ $y=1$ so we get
$$0=0+C$$
So $$y(t)=exp(int_0^tf(t)dt)$$
So either $2$ is true or $4$ is true? I am not able to go further from here. What logic should I use now
Edit:
Since $int_0^t f$ is continuous and $e^x$ is continuous and composition of continuous function is continuous so $2$ is true. Is this a correct way of thinking?
differential-equations initial-value-problems
asked 4 hours ago
StammeringMathematician
1,012217
Kind of an unusual multiple choice question when the choices are far from being mutually exclusive :)
– Erick Wong
46 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Consider the initial value problem $$y'(t)=f(t)y(t)$$ with $y(0)=1$ where $f:mathbbRrightarrow mathbbR$ is a continuous function.
Then this initial value problem has
(1) Infinitely many solutions for some $f$
(2)a unique solution in $mathbbR$
(3) no solution in $mathbbR$ for some $f$
(4) a solution in an interval containing $0$, but not on $mathbbR$ for some $f$
My efforts
$$fracdy(t)y(t) = f(t)dt$$
$$log y(t) = int_0^tf(t)dt +C$$
Now at $t=0$ $y=1$ so we get
$$0=0+C$$
So $$y(t)=exp(int_0^tf(t)dt)$$
So either $2$ is true or $4$ is true? I am not able to go further from here. What logic should I use now
Edit:
Since $int_0^t f$ is continuous and $e^x$ is continuous and composition of continuous function is continuous so $2$ is true. Is this a correct way of thinking?
differential-equations initial-value-problems
asked 4 hours ago
StammeringMathematician
1,012217
Consider the initial value problem $$y'(t)=f(t)y(t)$$ with $y(0)=1$ where $f:mathbbRrightarrow mathbbR$ is a continuous function.
Then this initial value problem has
(1) Infinitely many solutions for some $f$
(2)a unique solution in $mathbbR$
(3) no solution in $mathbbR$ for some $f$
(4) a solution in an interval containing $0$, but not on $mathbbR$ for some $f$
My efforts
$$fracdy(t)y(t) = f(t)dt$$
$$log y(t) = int_0^tf(t)dt +C$$
Now at $t=0$ $y=1$ so we get
$$0=0+C$$
So $$y(t)=exp(int_0^tf(t)dt)$$
So either $2$ is true or $4$ is true? I am not able to go further from here. What logic should I use now
Edit:
Since $int_0^t f$ is continuous and $e^x$ is continuous and composition of continuous function is continuous so $2$ is true. Is this a correct way of thinking?
differential-equations initial-value-problems
differential-equations initial-value-problems
asked 4 hours ago
StammeringMathematician
1,012217
asked 4 hours ago
StammeringMathematician
1,012217
edited 3 hours ago
asked 4 hours ago
StammeringMathematician
1,012217
asked 4 hours ago
StammeringMathematician
1,012217
asked 4 hours ago
StammeringMathematician
1,012217
1,012217
Kind of an unusual multiple choice question when the choices are far from being mutually exclusive :)
– Erick Wong
46 mins ago
add a comment |Â
Kind of an unusual multiple choice question when the choices are far from being mutually exclusive :)
– Erick Wong
46 mins ago
Kind of an unusual multiple choice question when the choices are far from being mutually exclusive :)
– Erick Wong
46 mins ago
Kind of an unusual multiple choice question when the choices are far from being mutually exclusive :)
– Erick Wong
46 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
I think the correct choice is (2) and only (2).
I reason as follows:
For $f(t) in C(Bbb R, Bbb R)$, consider the function
$F_f: Bbb R times Bbb R to Bbb R, ; F_f(y, t) = yf(t); tag 1$
then $F_f(y, t)$ is (a.) jointly continuous in $y$ and $t$, and (b.) (locally in $t$) Lipschitz continuous in $y$.
To see (a.), note that $F_f(y, t) = yf(t)$ is the product of the two continuous functions $y: Bbb R to Bbb R$ and $f: Bbb R to Bbb R$, and that multiplication is itself continuous from $Bbb R times Bbb R to Bbb R$.
As for (b.), we have, for $y, z, t in Bbb R$,
$vert F(z, t) - F(y, t) vert = vert zf(t) - yf(t) vert = vert(z - y)f(t) vert = vert z - y vert vert f(t) vert; tag 2$
now since $f: Bbb R to Bbb R$ is continuous, in any closed interval $I_t_0$ about any $t_0 in Bbb R$, $f(t)$ is bounded there; thus, there is a neighborhood $J_t_0 subset I_t_0$ of $t_0$ and a constant $L_J > 0$ such that
$vert f(t) vert < L_J, ; t in J_t_0; tag 3$
thus for $z, y in Bbb R$ and $t in J$ we have
$vert F(z, t) - F(y, t) vert < L_J vert z - y vert, tag 4$
which shows that $F(y, t)$ is Lipschitz continuous on some interval $J_t_0$ containing $t_0$; thus, by the Picard-Lindeloef theorem the exists a unique solution to our differential equation
$dot y = F(y, t) = f(t)y, ; y(t_0) = y_0, tag 5$
on some interval $(t_0 - epsilon, t_0 + epsilon)$ containing $t_0$; furthermore, since this is true for any $t_0$, such integral curves may be maximally extended to all of $Bbb R$ and thus we see there exists a unique, global solution passing through any $(y_0, t_0) in Bbb R times Bbb R$; since this is true for any $f(t) in C(Bbb R, Bbb R)$, we may rule out possibilities (1) and (3) in our OP Stammering Mathematician's list. Similarly, we may eliminate item (4) from consideration since, as we have seen, there is a solution for every $f$ in an interval about any $t_0$, $t_0 = 0$ included.
So, (2) is true and only (2) is true.
The particularly simple form of $F(y, t) = f(t)y$ in fact makes it possible to verify the above results in an immediate, practical way without resorting to the theoretial arguments presented above; indeed, if
$y(t_0) = y_0 ne 0 tag 6$
for some $t_0 in Bbb R$, then in a neighborhood of $t_0$ we have
$dfracdot y(t)y(t) = f(t), tag 7$
whence
$dfracdln y(t)dt = f(t), tag 8$
so that
$ln left (dfracy(t)y(t_0) right ) = ln y(t) - ln y(t_0) = displaystyle int_t_0^t dfracdln y(s)ds ; ds = int_t_0^t f(s) ; ds, tag 9$
and so
$y(t) = y(t_0) exp left (displaystyle int_t_0^t f(s) ; ds right ). tag10$
The solution (10) is easily seen to be valid, unique, and to extend to all $Bbb R$, and binds as long as $y(t_0) ne 0$ for some $t_0 in Bbb R$; the only other alternative is $y(t) = 0$ everywhere, which is also seen to be a unique global solution.
Finally, if we choose with our OP
$y(0) = 1, tag11$
we find the unique global solution to be
$y(t) = exp left ( displaystyle int_0^t f(s) ; ds right ), tag12$
in accord with our previous calculations.
answered 1 hour ago
Robert Lewis
39.3k22459
1
I don't know how to thank you. This is great explanation!!! Now I understand what is going on in the backdrop.. Thanks again for taking time to write such a long and detailed answer.
– StammeringMathematician
1 hour ago
1
@StammeringMathematician: my pleasure, my friend. Thanks for the kind words, and the "acceptance". Cheers!
– Robert Lewis
1 hour ago
add a comment |Â
up vote
2
down vote
You are correct. Since the integral, $int _0^t f(t)dt $ exists for all t, you have a solution for all t.
answered 3 hours ago
Mohammad Riazi-Kermani
32.7k41853
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
I think the correct choice is (2) and only (2).
I reason as follows:
For $f(t) in C(Bbb R, Bbb R)$, consider the function
$F_f: Bbb R times Bbb R to Bbb R, ; F_f(y, t) = yf(t); tag 1$
then $F_f(y, t)$ is (a.) jointly continuous in $y$ and $t$, and (b.) (locally in $t$) Lipschitz continuous in $y$.
To see (a.), note that $F_f(y, t) = yf(t)$ is the product of the two continuous functions $y: Bbb R to Bbb R$ and $f: Bbb R to Bbb R$, and that multiplication is itself continuous from $Bbb R times Bbb R to Bbb R$.
As for (b.), we have, for $y, z, t in Bbb R$,
$vert F(z, t) - F(y, t) vert = vert zf(t) - yf(t) vert = vert(z - y)f(t) vert = vert z - y vert vert f(t) vert; tag 2$
now since $f: Bbb R to Bbb R$ is continuous, in any closed interval $I_t_0$ about any $t_0 in Bbb R$, $f(t)$ is bounded there; thus, there is a neighborhood $J_t_0 subset I_t_0$ of $t_0$ and a constant $L_J > 0$ such that
$vert f(t) vert < L_J, ; t in J_t_0; tag 3$
thus for $z, y in Bbb R$ and $t in J$ we have
$vert F(z, t) - F(y, t) vert < L_J vert z - y vert, tag 4$
which shows that $F(y, t)$ is Lipschitz continuous on some interval $J_t_0$ containing $t_0$; thus, by the Picard-Lindeloef theorem the exists a unique solution to our differential equation
$dot y = F(y, t) = f(t)y, ; y(t_0) = y_0, tag 5$
on some interval $(t_0 - epsilon, t_0 + epsilon)$ containing $t_0$; furthermore, since this is true for any $t_0$, such integral curves may be maximally extended to all of $Bbb R$ and thus we see there exists a unique, global solution passing through any $(y_0, t_0) in Bbb R times Bbb R$; since this is true for any $f(t) in C(Bbb R, Bbb R)$, we may rule out possibilities (1) and (3) in our OP Stammering Mathematician's list. Similarly, we may eliminate item (4) from consideration since, as we have seen, there is a solution for every $f$ in an interval about any $t_0$, $t_0 = 0$ included.
So, (2) is true and only (2) is true.
The particularly simple form of $F(y, t) = f(t)y$ in fact makes it possible to verify the above results in an immediate, practical way without resorting to the theoretial arguments presented above; indeed, if
$y(t_0) = y_0 ne 0 tag 6$
for some $t_0 in Bbb R$, then in a neighborhood of $t_0$ we have
$dfracdot y(t)y(t) = f(t), tag 7$
whence
$dfracdln y(t)dt = f(t), tag 8$
so that
$ln left (dfracy(t)y(t_0) right ) = ln y(t) - ln y(t_0) = displaystyle int_t_0^t dfracdln y(s)ds ; ds = int_t_0^t f(s) ; ds, tag 9$
and so
$y(t) = y(t_0) exp left (displaystyle int_t_0^t f(s) ; ds right ). tag10$
The solution (10) is easily seen to be valid, unique, and to extend to all $Bbb R$, and binds as long as $y(t_0) ne 0$ for some $t_0 in Bbb R$; the only other alternative is $y(t) = 0$ everywhere, which is also seen to be a unique global solution.
Finally, if we choose with our OP
$y(0) = 1, tag11$
we find the unique global solution to be
$y(t) = exp left ( displaystyle int_0^t f(s) ; ds right ), tag12$
in accord with our previous calculations.
answered 1 hour ago
Robert Lewis
39.3k22459
1
I don't know how to thank you. This is great explanation!!! Now I understand what is going on in the backdrop.. Thanks again for taking time to write such a long and detailed answer.
– StammeringMathematician
1 hour ago
1
@StammeringMathematician: my pleasure, my friend. Thanks for the kind words, and the "acceptance". Cheers!
– Robert Lewis
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
I think the correct choice is (2) and only (2).
I reason as follows:
For $f(t) in C(Bbb R, Bbb R)$, consider the function
$F_f: Bbb R times Bbb R to Bbb R, ; F_f(y, t) = yf(t); tag 1$
then $F_f(y, t)$ is (a.) jointly continuous in $y$ and $t$, and (b.) (locally in $t$) Lipschitz continuous in $y$.
To see (a.), note that $F_f(y, t) = yf(t)$ is the product of the two continuous functions $y: Bbb R to Bbb R$ and $f: Bbb R to Bbb R$, and that multiplication is itself continuous from $Bbb R times Bbb R to Bbb R$.
As for (b.), we have, for $y, z, t in Bbb R$,
$vert F(z, t) - F(y, t) vert = vert zf(t) - yf(t) vert = vert(z - y)f(t) vert = vert z - y vert vert f(t) vert; tag 2$
now since $f: Bbb R to Bbb R$ is continuous, in any closed interval $I_t_0$ about any $t_0 in Bbb R$, $f(t)$ is bounded there; thus, there is a neighborhood $J_t_0 subset I_t_0$ of $t_0$ and a constant $L_J > 0$ such that
$vert f(t) vert < L_J, ; t in J_t_0; tag 3$
thus for $z, y in Bbb R$ and $t in J$ we have
$vert F(z, t) - F(y, t) vert < L_J vert z - y vert, tag 4$
which shows that $F(y, t)$ is Lipschitz continuous on some interval $J_t_0$ containing $t_0$; thus, by the Picard-Lindeloef theorem the exists a unique solution to our differential equation
$dot y = F(y, t) = f(t)y, ; y(t_0) = y_0, tag 5$
on some interval $(t_0 - epsilon, t_0 + epsilon)$ containing $t_0$; furthermore, since this is true for any $t_0$, such integral curves may be maximally extended to all of $Bbb R$ and thus we see there exists a unique, global solution passing through any $(y_0, t_0) in Bbb R times Bbb R$; since this is true for any $f(t) in C(Bbb R, Bbb R)$, we may rule out possibilities (1) and (3) in our OP Stammering Mathematician's list. Similarly, we may eliminate item (4) from consideration since, as we have seen, there is a solution for every $f$ in an interval about any $t_0$, $t_0 = 0$ included.
So, (2) is true and only (2) is true.
The particularly simple form of $F(y, t) = f(t)y$ in fact makes it possible to verify the above results in an immediate, practical way without resorting to the theoretial arguments presented above; indeed, if
$y(t_0) = y_0 ne 0 tag 6$
for some $t_0 in Bbb R$, then in a neighborhood of $t_0$ we have
$dfracdot y(t)y(t) = f(t), tag 7$
whence
$dfracdln y(t)dt = f(t), tag 8$
so that
$ln left (dfracy(t)y(t_0) right ) = ln y(t) - ln y(t_0) = displaystyle int_t_0^t dfracdln y(s)ds ; ds = int_t_0^t f(s) ; ds, tag 9$
and so
$y(t) = y(t_0) exp left (displaystyle int_t_0^t f(s) ; ds right ). tag10$
The solution (10) is easily seen to be valid, unique, and to extend to all $Bbb R$, and binds as long as $y(t_0) ne 0$ for some $t_0 in Bbb R$; the only other alternative is $y(t) = 0$ everywhere, which is also seen to be a unique global solution.
Finally, if we choose with our OP
$y(0) = 1, tag11$
we find the unique global solution to be
$y(t) = exp left ( displaystyle int_0^t f(s) ; ds right ), tag12$
in accord with our previous calculations.
answered 1 hour ago
Robert Lewis
39.3k22459
1
I don't know how to thank you. This is great explanation!!! Now I understand what is going on in the backdrop.. Thanks again for taking time to write such a long and detailed answer.
– StammeringMathematician
1 hour ago
1
@StammeringMathematician: my pleasure, my friend. Thanks for the kind words, and the "acceptance". Cheers!
– Robert Lewis
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
I think the correct choice is (2) and only (2).
I reason as follows:
For $f(t) in C(Bbb R, Bbb R)$, consider the function
$F_f: Bbb R times Bbb R to Bbb R, ; F_f(y, t) = yf(t); tag 1$
then $F_f(y, t)$ is (a.) jointly continuous in $y$ and $t$, and (b.) (locally in $t$) Lipschitz continuous in $y$.
To see (a.), note that $F_f(y, t) = yf(t)$ is the product of the two continuous functions $y: Bbb R to Bbb R$ and $f: Bbb R to Bbb R$, and that multiplication is itself continuous from $Bbb R times Bbb R to Bbb R$.
As for (b.), we have, for $y, z, t in Bbb R$,
$vert F(z, t) - F(y, t) vert = vert zf(t) - yf(t) vert = vert(z - y)f(t) vert = vert z - y vert vert f(t) vert; tag 2$
now since $f: Bbb R to Bbb R$ is continuous, in any closed interval $I_t_0$ about any $t_0 in Bbb R$, $f(t)$ is bounded there; thus, there is a neighborhood $J_t_0 subset I_t_0$ of $t_0$ and a constant $L_J > 0$ such that
$vert f(t) vert < L_J, ; t in J_t_0; tag 3$
thus for $z, y in Bbb R$ and $t in J$ we have
$vert F(z, t) - F(y, t) vert < L_J vert z - y vert, tag 4$
which shows that $F(y, t)$ is Lipschitz continuous on some interval $J_t_0$ containing $t_0$; thus, by the Picard-Lindeloef theorem the exists a unique solution to our differential equation
$dot y = F(y, t) = f(t)y, ; y(t_0) = y_0, tag 5$
on some interval $(t_0 - epsilon, t_0 + epsilon)$ containing $t_0$; furthermore, since this is true for any $t_0$, such integral curves may be maximally extended to all of $Bbb R$ and thus we see there exists a unique, global solution passing through any $(y_0, t_0) in Bbb R times Bbb R$; since this is true for any $f(t) in C(Bbb R, Bbb R)$, we may rule out possibilities (1) and (3) in our OP Stammering Mathematician's list. Similarly, we may eliminate item (4) from consideration since, as we have seen, there is a solution for every $f$ in an interval about any $t_0$, $t_0 = 0$ included.
So, (2) is true and only (2) is true.
The particularly simple form of $F(y, t) = f(t)y$ in fact makes it possible to verify the above results in an immediate, practical way without resorting to the theoretial arguments presented above; indeed, if
$y(t_0) = y_0 ne 0 tag 6$
for some $t_0 in Bbb R$, then in a neighborhood of $t_0$ we have
$dfracdot y(t)y(t) = f(t), tag 7$
whence
$dfracdln y(t)dt = f(t), tag 8$
so that
$ln left (dfracy(t)y(t_0) right ) = ln y(t) - ln y(t_0) = displaystyle int_t_0^t dfracdln y(s)ds ; ds = int_t_0^t f(s) ; ds, tag 9$
and so
$y(t) = y(t_0) exp left (displaystyle int_t_0^t f(s) ; ds right ). tag10$
The solution (10) is easily seen to be valid, unique, and to extend to all $Bbb R$, and binds as long as $y(t_0) ne 0$ for some $t_0 in Bbb R$; the only other alternative is $y(t) = 0$ everywhere, which is also seen to be a unique global solution.
Finally, if we choose with our OP
$y(0) = 1, tag11$
we find the unique global solution to be
$y(t) = exp left ( displaystyle int_0^t f(s) ; ds right ), tag12$
in accord with our previous calculations.
answered 1 hour ago
Robert Lewis
39.3k22459
I think the correct choice is (2) and only (2).
I reason as follows:
For $f(t) in C(Bbb R, Bbb R)$, consider the function
$F_f: Bbb R times Bbb R to Bbb R, ; F_f(y, t) = yf(t); tag 1$
then $F_f(y, t)$ is (a.) jointly continuous in $y$ and $t$, and (b.) (locally in $t$) Lipschitz continuous in $y$.
To see (a.), note that $F_f(y, t) = yf(t)$ is the product of the two continuous functions $y: Bbb R to Bbb R$ and $f: Bbb R to Bbb R$, and that multiplication is itself continuous from $Bbb R times Bbb R to Bbb R$.
As for (b.), we have, for $y, z, t in Bbb R$,
$vert F(z, t) - F(y, t) vert = vert zf(t) - yf(t) vert = vert(z - y)f(t) vert = vert z - y vert vert f(t) vert; tag 2$
now since $f: Bbb R to Bbb R$ is continuous, in any closed interval $I_t_0$ about any $t_0 in Bbb R$, $f(t)$ is bounded there; thus, there is a neighborhood $J_t_0 subset I_t_0$ of $t_0$ and a constant $L_J > 0$ such that
$vert f(t) vert < L_J, ; t in J_t_0; tag 3$
thus for $z, y in Bbb R$ and $t in J$ we have
$vert F(z, t) - F(y, t) vert < L_J vert z - y vert, tag 4$
which shows that $F(y, t)$ is Lipschitz continuous on some interval $J_t_0$ containing $t_0$; thus, by the Picard-Lindeloef theorem the exists a unique solution to our differential equation
$dot y = F(y, t) = f(t)y, ; y(t_0) = y_0, tag 5$
on some interval $(t_0 - epsilon, t_0 + epsilon)$ containing $t_0$; furthermore, since this is true for any $t_0$, such integral curves may be maximally extended to all of $Bbb R$ and thus we see there exists a unique, global solution passing through any $(y_0, t_0) in Bbb R times Bbb R$; since this is true for any $f(t) in C(Bbb R, Bbb R)$, we may rule out possibilities (1) and (3) in our OP Stammering Mathematician's list. Similarly, we may eliminate item (4) from consideration since, as we have seen, there is a solution for every $f$ in an interval about any $t_0$, $t_0 = 0$ included.
So, (2) is true and only (2) is true.
The particularly simple form of $F(y, t) = f(t)y$ in fact makes it possible to verify the above results in an immediate, practical way without resorting to the theoretial arguments presented above; indeed, if
$y(t_0) = y_0 ne 0 tag 6$
for some $t_0 in Bbb R$, then in a neighborhood of $t_0$ we have
$dfracdot y(t)y(t) = f(t), tag 7$
whence
$dfracdln y(t)dt = f(t), tag 8$
so that
$ln left (dfracy(t)y(t_0) right ) = ln y(t) - ln y(t_0) = displaystyle int_t_0^t dfracdln y(s)ds ; ds = int_t_0^t f(s) ; ds, tag 9$
and so
$y(t) = y(t_0) exp left (displaystyle int_t_0^t f(s) ; ds right ). tag10$
The solution (10) is easily seen to be valid, unique, and to extend to all $Bbb R$, and binds as long as $y(t_0) ne 0$ for some $t_0 in Bbb R$; the only other alternative is $y(t) = 0$ everywhere, which is also seen to be a unique global solution.
Finally, if we choose with our OP
$y(0) = 1, tag11$
we find the unique global solution to be
$y(t) = exp left ( displaystyle int_0^t f(s) ; ds right ), tag12$
in accord with our previous calculations.
answered 1 hour ago
Robert Lewis
39.3k22459
edited 1 hour ago
answered 1 hour ago
Robert Lewis
39.3k22459
answered 1 hour ago
Robert Lewis
39.3k22459
answered 1 hour ago
Robert Lewis
39.3k22459
39.3k22459
1
I don't know how to thank you. This is great explanation!!! Now I understand what is going on in the backdrop.. Thanks again for taking time to write such a long and detailed answer.
– StammeringMathematician
1 hour ago
1
@StammeringMathematician: my pleasure, my friend. Thanks for the kind words, and the "acceptance". Cheers!
– Robert Lewis
1 hour ago
add a comment |Â
1
I don't know how to thank you. This is great explanation!!! Now I understand what is going on in the backdrop.. Thanks again for taking time to write such a long and detailed answer.
– StammeringMathematician
1 hour ago
1
@StammeringMathematician: my pleasure, my friend. Thanks for the kind words, and the "acceptance". Cheers!
– Robert Lewis
1 hour ago
1
1
I don't know how to thank you. This is great explanation!!! Now I understand what is going on in the backdrop.. Thanks again for taking time to write such a long and detailed answer.
– StammeringMathematician
1 hour ago
I don't know how to thank you. This is great explanation!!! Now I understand what is going on in the backdrop.. Thanks again for taking time to write such a long and detailed answer.
– StammeringMathematician
1 hour ago
1
1
@StammeringMathematician: my pleasure, my friend. Thanks for the kind words, and the "acceptance". Cheers!
– Robert Lewis
1 hour ago
@StammeringMathematician: my pleasure, my friend. Thanks for the kind words, and the "acceptance". Cheers!
– Robert Lewis
1 hour ago
add a comment |Â
up vote
2
down vote
You are correct. Since the integral, $int _0^t f(t)dt $ exists for all t, you have a solution for all t.
answered 3 hours ago
Mohammad Riazi-Kermani
32.7k41853
add a comment |Â
up vote
2
down vote
You are correct. Since the integral, $int _0^t f(t)dt $ exists for all t, you have a solution for all t.
answered 3 hours ago
Mohammad Riazi-Kermani
32.7k41853
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You are correct. Since the integral, $int _0^t f(t)dt $ exists for all t, you have a solution for all t.
answered 3 hours ago
Mohammad Riazi-Kermani
32.7k41853
You are correct. Since the integral, $int _0^t f(t)dt $ exists for all t, you have a solution for all t.
answered 3 hours ago
Mohammad Riazi-Kermani
32.7k41853
answered 3 hours ago
Mohammad Riazi-Kermani
32.7k41853
answered 3 hours ago
Mohammad Riazi-Kermani
32.7k41853
answered 3 hours ago
Mohammad Riazi-Kermani
32.7k41853
32.7k41853
add a comment |Â
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%2f2934940%2fa-multiple-choice-question-concerning-first-order-linear-differential-equations%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
Kind of an unusual multiple choice question when the choices are far from being mutually exclusive :)
– Erick Wong
46 mins ago